I claim all such functions are constant functions: there is a $c\in\mathbb{R}$ such that $f(x)=g(y)=c$ for all $x,y>0$. To that end, I first show $f(a)=g(b)$ for every $a\ge 2b$. For that, it suffices to show the system, $a=x^2+y^2$ and $b=xy$, admits a solution $x,y>0$ for every $a\ge 2b$. Setting $y=b/x$, it boils down \[ a=x^2 + \frac{b^2}{x^2} = \frac{x^4+b^2}{x^2}\iff x^4 - ax^2 + b^2 = 0. \]Clearly if $a\ge 2b$, \[ (x,y)=\left(\sqrt{\frac{a+\sqrt{a^2-4b^2}}{2}},\sqrt{\frac{a-\sqrt{a^2-4b^2}}{2}}\right). \]Having established the claim, the rest is easy. Assume $g$ is not constant: for some $b_1<b_2$, $g(b_1)=g(b_2)$. Setting $a=2b_2$, we find $f(a)=g(b_2)$ and $f(a)=g(b_1)$, a clear contradiction. From here, $g$ is constant and so is $f$.

We will prove that both $f,g$ are constant and equal. Let $x \rightarrow kx_{1} , y \rightarrow y_{1}/k$, for some $k>0$. It occurs that $g(x_{1}y_{1}) = f(k^2x_{1}^2 + y_{1}^2/k^2) \Rightarrow f(x_{1}^2 + y_{1}^2) = f(k^2x_{1}^2 + y_{1}^2/k^2)$, for all $k >0$ It can easily be proven that the equation $k^2x_{1}^2 + y_{1}^2/k^2 = c$ with unknown $k$ has at least one positive root for all $c\geq2x_{1}y_{1}$, and thus for all $c\geq x_{1}^2 + y_{1}^2$. Therefore $f(x)$ is constant for all $x\geq x_{1}^2 + y_{1}^2$ Similarly , for all $x_{0}, y_{0}$ such that $x_{0}^2 + y_{0}^2 < x_{1}^2 + y_{1}^2$ exists $k_{0}$ such that $k_{0}^2x_{0}^2 + y_{0}^2/k_{0}^2 = x_{1}^2 + y_{1}^2 \Rightarrow f(x_{0}^2 + y_{0}^2) = f(x_{1}^2 + y_{1}^2)$ since $x_{1}^2 + y_{1}^2 > x_{0}^2 + y_{0}^2 \geq 2x_{0}y_{0}$. Therefore $f(x)$ is constant for all $0 < x\leq x_{1}^2 + y_{1}^2$. Thus $f(x)$ is constant , so is $g(x)$, as desired

Let $c>0$ be arbitrary and set $y=\frac{c}{x}$. Then $f(x^2 + \frac{c^2}{x^2}) = g(c)$ and since $x^2 + \frac{c^2}{x^2} \geq 2c$ runs through the interval $[2c,\infty)$, we get that $f(x)$ is constant on $[2c,\infty)$. As $c$ was arbitrary, we conclude that $f$ is constant. Finally, $y=1$ shows that $f\equiv g$ are the same constant (and conversely, all such work).

Let $y = \frac{a}{x}$ then we have $f(x^2 + \frac{a^2}{x^2}) = g(a)$ so for every $a,b$ such that $b \ge 2a$ we have $f(b) = g(a)$. Assume $f$ is not constant so there exists $t < k$ such that $f(t) \neq f(k)$ but then for $a = \frac{t}{2}$ we have $f(t) = f(k)$ so contradiction so $f$ is constant. Now assume that $g$ is not constant then there exists $t < k$ such that $g(t) \neq g(k)$ but then let $b = 2k$ so now $b \ge 2k$ and $b > 2t$ so $f(b) = g(k)$ and $f(b) = g(t)$ so contradiction so $g$ is constant.

Another way to motivate the solution is by setting \((x,y) = (\sin(t), \cos(t))\).

Let $P(x;y)$ denotes the assertion of given functional equation. Note that $P(\sqrt2x; \sqrt2y)$ reveals the following: $$g(2xy)= f(2x^2 +2y^2) = f((x+y)^2+(x-y)^2) = g((x-y)(x+y)) =g(x^2-y^2) $$for real numbers $x,y$ obeying $x>y$. Suppose $x>1$, then $x>\frac{1}{x}$. Thus replacing $y$ by $\frac{1}{x}$ into beforehand established result yields to: $$ g(2) = g(x^2 - \frac{1}{x^2})$$. It remains to prove that function $h(x) = x^2 -\frac{1}{x^2}$ is surjective over positive real numbers for $x>1$. This is equivalent to showing that the equation: $$ x^2 - \frac{1}{x^2} =a \implies x^4 -ax^2 - 1 = 0 $$has always a real root greater than $1$ for every real number $a>0$. Indeed just note that $x = \sqrt{\frac{a+\sqrt{a^2+4}}{2}}$ is always a root and $x = \sqrt{\frac{a+\sqrt{a^2+4}}{2}} >\sqrt{\frac{0+\sqrt{0+4}}{2}}=1$ as needed. This allows us to conclude that $g(x)$ is constant, which immediately implies that $f(x)$ is the same constant too, since $x^2+y^2$ varies through all positive real numbers as $x,y$ vary.

We will show that both $f$ and $g$ are constant functions with the same value. The solution clearly works and we will now prove that they are the only ones. Denote the assertion the Functional Equation as $P(x,y)$. From $P\left(x,\frac{1}{x}\right)$ we get $f\left(x^2+\frac{1}{x^2}\right)= g(1)$. By AM-GM Inequality, clearly $x^2+\frac{1}{x^2} \ge 2$. This means $f(x)=g(1)$ for all $x \ge 2$. With $P\left(\frac{x}{2},2\right)$ we get \[g(x) = f\left(\frac{x^2}{4} + 4\right)\]This shows that $g(x)$ is always constant because $\frac{x^2}{4} + 4 > 4 > 2$. Now, as $g$ is constant for every $x$ so must $f(x)$ and thus we are done. $\blacksquare$

Let $x= c \cdot \sin(t), y= c \cdot \cos (t)$ for positive real numbers $x,y,t$. We have: $$f(c^2)= g(c^2 \cos(t)\sin(t))$$ Observe that $LHS$ is a constant , therefore $g$ is constant for domain $c^2 \cos(t) \sin(t)$. But since $c$ is arbitrary, we can conclude that for domain $\mathbb{R}^{+}$, functions $f,g$ are constant functions. Letting $x=1$, we can conclude $f \equiv g \ \blacksquare$

Let $a$ be some parameter, for which we will make $y=\tfrac{a}{x}$. Then, $f\left(x^2+\tfrac{a}{x^2}\right)=g(a)$ The range of $x^2+\tfrac{a}{x^2}$ under the domain $x\in (0,\infty)$ is $[2a,\infty)$. Therefore, if we make $a$ sufficiently small then we find $f$ to be constant. Now, if we vary $a$ instead of $x$, we find that $g(a)$ is constant. Therefore, $f,g$ are equal constant functions, and these solutions obviously work.

We claim the only such functions are of the form $f(x) = g(x) = c$ where $c$ is a positive real constant. It is easily visible that this pair satisfies the equation. Now we prove it is the only such function. Note that $f(2x^2) = g(x^2) \implies f(2x) = g(x) \implies g(xy) = f(2xy) \implies f(x^2 + y^2) = f(2xy)$. Now see that we can consider positive reals $a \ge b$ which satisfy $a = x^2 + y^2, b = 2xy$. ($a\ge b$ comes from the AM-GM inequality.) This gives us $x =\frac{ \sqrt{a+b} + \sqrt{a-b}}{2}, y = \frac{ \sqrt{a+b} - \sqrt{a-b}}{2}$, which indeed works satisfies the way we have defined $a$ and $b$. Thus, $f(a) = f(b)$ for all $a$ and $b$, implying $f$ is constant, and as $g(x) = f(2x)$, $g$ is the same constant, QED.

not quite sure if this approach works: let $P(x,y)$ denote the functional equation. $P(x,\frac{1}{x})$ gives us that if $g(1)=c$ for some constant $c$, then $f(a) = c$ for $a\in \left[ 2, \infty \right)$, this follows trivially by AM-GM Now we have proven that $f(x)$ eventually becomes constant. Now let $x=2$ and let $y$ vary over the positive real numbers. clearly, $2y$ achieves any value in the positive real numbers, but $2^2 + \frac{1}{y^2} \geq 2$ for any $y$, which leads to the conclusion that $g(x) = c$ for any positive real number $x$ Now it remains to prove that if $g(x) = c$, that $f(x)$ is also constant, but this is obvious since $x^2+y^2$ can achieve any positive real value, and $g(xy)=c$ for any values of $x,y$. Since the both are equal we conclude that $f(x)=g(x)=c$ for any constant $c$

Plug in $x=x,y=1/x$ to get $f(x^2+1/x^2)=g(1)$ and $x^2+1/x^2\ge 2$ so $f$ is constant for $f(t),t\ge 2$. $y=x \implies f(2x^2)=g(x^2) \implies f(2x)=g(x)$ so $g$ is constant for $g(t),t \ge 1$.Just put $x \ge 2$and $y$ a small number to get $g$ is constant and this implies $f$ is constant as well,therefore $f(x)=g(x)=c$ for all $x$ is our answer.

If we set $x=y=\sqrt{r}$, for some positive real $r$, we see that $f(2r)=g(r)$. Hence, we have that $f(x^2+y^2) = f(2xy)$. Let $y = \tfrac{r}{x}$, for some positive real $x$; then, we have that $f(x^2+\tfrac{r^2}{x^2}) = f(2r)$. Note that by AM-GM, $x^2+\tfrac{r^2}{x^2}$ can take on any real value greater than or equal to $2r$, simply by varying $x$. This means that for any value of $2r$, whose range is the positive reals, then $f$ is constant on $[2r, \infty)$. Thus, $f$ is constant. The solutions are, $f\equiv c$, for some real constant $c$, and it follows that $g\equiv c$.