Solution: $ f(x)=c$ where $ c\in\mathbb{N}_{0}$ for any $ x\in\mathbb{N}_{0}$ First if we plug in $ m=0$ and $ n=1$ we get $ f(0)=f(1)$ Now, assume there exist numbers $ a,b\in\mathbb{N}_{0}$ not equal and neither equal to $ 0$ such that $ f(a)$ is not equal to $ f(b)$ wolog $ f(a)>f(b)$. If we plug in $ a=m$ and $ b=n$ in the original equation we get: $ af(b)+bf(a)=(a+b)f(a^{2}+b^{2})$. However, as $ f(a)>f(b)$ we have $ (a+b)f(b)<af(b)+bf(a)<(a+b)f(a)$. Then, $ (a+b)f(b)<(a+b)f(a^{2}+b^{2})<(a+b)f(a)$, and diving through by $ a+b$ we get: $ f(b)<f(a^{2}+b^{2})<f(a)$ but then we can just repeat this procedure with $ a$ and $ a^{2}+b^{2}$ to get another number $ c=(a^{2}+b^{2})^{2}+a^{2}$ so that $ f(b)<f(x)<f(a^{2}+b^{2})$ and we can repeat this forever and because the function is $ \mathbb{N}_{0}\to\mathbb{N}_{0}$ then we get infinitely many natural numbers in the range $ (f(b),f(a)$ which is impossible. Therefore we can never have two numbers $ a,b\in\mathbb{N}_{0}$ not equal and neither equal to $ 0$ so that $ f(a)$ is not equal to $ f(b)$. So, $ f(x)$ is constant, QED.

My solution From the functional equation we have: $0f(n)+nf(0)=nf(n^2)$ and so $f(n^2)=f(0)$ for all $n$ $3n^2f(4n^2)+4n^2f(3n^2)=7n^2f(25n^4)$ $\implies 3n^2f(0)+4n^2f(3n^2)=7n^2f(0)$ $\implies f(3n^2)=f(0)$ for all $n$ $12n^2f(5n^2)+5n^2f(12n^2)=17n^2f(169n^4)$ $\implies 12n^2f(5n^2)+5n^2f(0)=17n^2f(0)$ $\implies f(5n^2)=f(0)$ for all $n$ $2n^2f(n^2)+n^2f(2n^2)=3n^2f(5n^4)$ $\implies 2n^2f(0)+n^2f(2n^2)=3n^2f(0)$ $\implies f(2n^2)=f(0)$ for all $n$ $2nf(n)=nf(n)+nf(n)=2nf(2n^2)=2nf(0)$ $\implies f(n)=f(0)$ for all $n$ So $f$ is constant, which works.

Let's set n=0 --> $mf(0)=mf(m^2)$ --> for all m $f(m^2)=f(0)=K$ $m+n|mf(n)+nf(m)$ also $m+n|(m+n)f(m)$ --> $m+n|m(f(m)-f(n))$ now let's take n such that $n=t^2$ & $(m,n)=1$ --> $m+n|f(m)-f(n)$ we had $f(m^2)=f(0)=K$ --> $m+n|f(m)-K$ The left hand side can get as large as we need it to be but the right hand side will remain the same therefore the righthand side should be 0. --> The solution is $f(m)=K$

rem wrote: but then we can just repeat this procedure with $ a$ and $ a^{2}+b^{2}$ to get another number $ c=(a^{2}+b^{2})^{2}+a^{2}$ so that $ f(b)<f(x)<f(a^{2}+b^{2})$ Hello,rem!I think $c=(a^2+b^2)^2+b^2$ and $f(b)<f(c)<f(a^2+b^2)$. Thanks.