First, note that $(xy-1)(x+y)>0$. In particular, if $xy>1$, then we must necessarily have $x+y>0$, hence, $x,y>0$. Assuming $xy>1$, observe that, $$ x^2+y^2+x+y = xy(x+y)-\frac{10}{27}\leq \frac{25}{9}(x+y)-\frac{10}{27}. $$Rearranging this, we arrive at $48(x+y)\geq 27(x^2+y^2)+10$. Now, by Cauchy-Schwarz inequality we have $x^2+y^2\geq \frac{(x+y)^2}{2}$, hence, we have $$ 96(x+y)\geq 27(x+y)^2 + 20 \implies (9t-2)(3t-10)\leq 0 $$for $t=x+y$. In particular, $\frac29 \leq x+y\leq \frac{10}{3}$. \noindent Next, we lower bound $x^2+y^2+x+y\geq 2xy + 2\sqrt{xy}$. Hence, $$ x+y\geq 2+\frac{10}{27xy}+\frac{2}{\sqrt{xy}}. $$Next, using $xy\leq \frac{25}{9}$, we get $$ x+y\geq 2+\frac{10}{27\cdot \frac{25}{9}}+\frac{2}{\frac{5}{3}} = 10/3. $$Hence, we must have an equality throughout, and thus $x+y=\frac{10}{3}$, and $xy=\frac{25}{9}$ must hold. Now, plugging this into the original equation, we get $2(x^2+y^2)=\frac{100}{9}=(x+y)^2$, giving $(x-y)^2=0$. Hence, under $xy>1$ assumption, we get $(x,y)=(5/3,5/3)$ is the only solution. Next, suppose that $xy<1$. In particular, we have $x+y<0$. Assume that both $x,y<0$, and let $x=-a$ and $y=-b$. We can rewrite the equation as $0<ab<1$, and $a^2+b^2+ab(a+b)=a+b-\frac{10}{27}$. Now, letting $t=a+b$, and using $a^2+b^2\geq \frac{t^2}{2}$, we get, $$ t\geq \frac{t^2}{2}+\frac{10}{27}+abt \implies 0\geq \frac{t^2}{2}+(ab-1)t+\frac{10}{27}. $$Now, using AM-GM, we also have $t\geq 2\sqrt{ab}$, and letting $q=\sqrt{ab}$, this inequality can be rewritten as $(3q-1)^2(3q+5)\leq 0$. In particular, the only possibility is $q=\frac{1}{3}$, namely, $xy=ab=\frac{1}{9}$, and $a=b$. Hence, $(x,y)=(-1/3,-1/3)$ is the solution for this case. Finally, suppose that one of the $x,y$ is positive, and the other one is negative. Without loss, let $x=a$ and $y=-b$. We have, $$ a^2+b^2+a-b=-ab(a-b)-\frac{10}{27}\implies a^2+b^2+ab(a-b)+(a-b)=-\frac{10}{27}. $$Here, the first observation is $a-b<0$. Next, subtract $2ab$ from both sides, and arrive at, $$ (b-a)^2=(b-a)+(b-a-2)ab - \frac{10}{27}. $$Let $t=b-a$. If $b-a\geq 2$, then, $$ t^2 \leq t+(t-2)\frac{25}{9}-\frac{10}{27}, $$using $ab\leq \frac{25}{9}$, which yields a contradiction. Hence, $t<2$. In this case, note that $(b-a-2)ab<0$, which yields, $$ t^2\leq t-\frac{10}{27}, $$which, again has no solutions. Hence, the only solutions to the given equation are $\boxed{(x,y)=(5/3,5/3),(-1/3,-1/3)}$.

We define $a=xy,b=x+y$ so we have that $b^2\ge4a$ and $b\neq2$ So we have $a \leq \frac{b^2}{4}$ According to the problem statement we have that $b^2-2a+b=ab-\frac{10}{27}$ and $-\frac{25}{9} \leq a \leq \frac{25}{9}$ Simplifying the first statement gives us $a=\frac{b^2+b+\frac{27}{10}}{b+2}$

So, in order to solve the range of $b$, we only need to solve the system of inequalities below: $$\begin{cases} -\frac{25}{9} \leq \frac{b^2+b+\frac{27}{10}}{b+2} \leq \frac{25}{9} \\ \frac{b^2+b+\frac{27}{10}}{b+2} \leq \frac{b^2}{4} \end{cases}$$

First, by $b^2-2a+b=ab-\frac{10}{27}$, it's easy to verify that $b \neq -2$ For the First Inequality: We classify it into $b>-2$ and $b<-2$ and solve them independently. We can have that $-\frac{14}{9} \leq b \leq \frac{10}{3}$ For the Second Inequality:The same as above, we have that $b \ge \frac{10}{3} or b=-\frac{2}{3} or b<-2$ Combining the two above we have that $b=\frac{10}{3} or b=-\frac{2}{3}$

if $b=\frac{10}{3}$ we have that $a=\frac{25}{9}$ so $(x,y)=\left(\frac{5}{3}, \frac {5}{3}\right)$ if $b=-\frac{2}{3}$ we have that $a=\frac{1}{9}$ so $(x,y)=\left(-\frac{1}{3}, -\frac {1}{3}\right)$ We are done