I will show that $\boxed{(a,b,c,d,e,f)=(0,0,0,0,0,0)}$ is the only solution. Let $f'=abcde-f$ and earn $$2abcdef' = a^2+b^2+c^2+d^2+e^2+f'^2,$$which is a symmetric equation. See that if one is $0$, then all of them is $0$. Now let a set of solutions $S \subset \mathbb{N}^6$ such that $$S = \{ (a,b,c,d,e,f) | a,b,c,d,e,f \subset \mathbb{N}, a^2+b^2+c^2+d^2+e^2+f^2 = 2abcdef \neq 0 \}$$Assume that $S \neq \phi$. Now exist a solution that $(x_1, x_2, x_3, x_4, x_5,x_6) \in S$ such that $x_1+x_2+x_3+x_4+x_5+x_6$ has its minimum value. WLOG, $x_1 \geq x_2 \geq x_3 \geq x_4 \geq x_5 \geq x_6$. Let a quadratic equation $$f(x)=x^2-2x_2x_3x_4x_5x_6x+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2$$And since $x_1$ is a solution, so there is another solution $y$. - $y=\frac{x_2^2+x_3^2+x_4^2+x_5^2+x_6^2}{x_1}$, so $y \geq 0$ - $y=2x_2x_3x_4x_5x_6-x_1$, so $y \in \mathbb{Z}$ $\therefore y \in \mathbb{N}$. If $y < x_1$, $(y, x_2, x_3, x_4, x_5,x_6)$ is also a solution, and $x_1+x_2+x_3+x_4+x_5+x_6 > y+x_2+x_3+x_4+x_5+x_6$ contradicts the assumation to minimality of $(x_1, x_2, x_3, x_4, x_5,x_6)$. $\therefore y \geq x_1$, So $y \geq x_1 \geq x_2$. Since $x_1$ and $y$ are solutions to $f(t)=0$, $$f(t)=(t-x)(t-y).$$Also, $y \geq x_1 \geq x_2$, $f(x_2)=(x_2-x)(x_2-y) \geq 0$. $\therefore x_2^2-(2x_2x_3x_4x_5x_6)x_2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2 = f(x_2) \geq 0$, $$6x_2^2 \geq 2x_2^2+x_3^2+x_4^2+x_5^2+x_6^2 \geq (2x_2x_3x_4x_5x_6)x_2,$$$\therefore 3 \geq x_3x_4x_5x_6$. Case $1$) $(x_3,x_4,x_5,x_6)=(3,1,1,1)$ We get $x_1^2+x_2^2+12=6x_1x_2$. By$\mod 3$ we earn $3 | x_1, x_2$, which contradicts$\mod 9$. Case $2$) $(x_3,x_4,x_5,x_6)=(2,1,1,1)$ We get $x_1^2+x_2^2+7=4x_1x_2$. We do the original vieta jumping again. Let a set of solutions $T \subset \mathbb{N}^2$ such that $$T = \{ (x_1, x_2) | x_1, x_2 \subset \mathbb{N}, x_1^2+x_2^2+7=4x_1x_2 \}$$and let a solution $(s,t)$ such that $s+t$ has the minimun value. Let a quadratic equation $$g(x)=x^2-4tx+t^2+7$$and let the solution set of $g(x)=0$ be $\{ s, s' \}$. If $s' < s$, then $(s', t) \in T$, which is a contradiction. $\therefore s' \geq s$, $s' \geq s \geq t$. So $$t^2-4t^2+t^2+7=g(t)=(t-s)(t-s') \geq 0$$$$7 \geq 2t^2$$$$t=1$$Then $s^2-4s+8=0$, which is a contradiction. Case $3$) $(x_3,x_4,x_5,x_6)=(1,1,1,1)$ We get $x_1^2+x_2^2+4=2x_1x_2$. By $(x_1-x_2)^2+4=0$, it's a contradiction. Therefore $S = \phi$, and we conclude that $(a,b,c,d,e,f')=(0,0,0,0,0,0)$, which we earn $\boxed{(a,b,c,d,e,f)=(0,0,0,0,0,0)}$, as desired.