Solution:
We claim that $f$ is a constant function. Suppose, for a contradiction, that there exist $x$ and $y$ with $f(x) < f(y)$; choose $x, y$ such that $f(y) - f(x) > 0$ is minimal. Then
$$f(x) = \frac{x f(x) + y f(x)}{x + y} < \frac{x f(y) + y f(x)}{x + y} < \frac{x f(y) + y f(y)}{x + y} = f(y)$$
so $f(x) < f\left(x^{2} + y^{2}\right) < f(y)$ and $0 < f\left(x^{2} + y^{2}\right) - f(x) < f(y) - f(x)$, contradicting the choice of $x$ and $y$. Thus, $f$ is a constant function. Since $f(0)$ is in $\mathbb{N}$, the constant must be from $\mathbb{N}$.
Also, for any $c$ in $\mathbb{N}$, $x c + y c = (x + y) c$ for all $x$ and $y$, so $f(x) = c$, $c \in \mathbb{N}$ are the solutions to the equation.

We claim $f$ is a constant function. Define $g(x) = f(x) - f(0)$. Then $g(0) = 0$, $g(x) \geq -f(0)$ and
$$x g(y) + y g(x) = (x + y) g\left(x^{2} + y^{2}\right)$$
for all $x, y$ in $\mathbb{N}$.
Letting $y = 0$ shows $g\left(x^{2}\right) = 0$ (in particular, $g(1) = g(4) = 0$), and letting $x = y = 1$ shows $g(2) = 0$. Also, if $x, y$ and $z$ in $\mathbb{N}$ satisfy $x^{2} + y^{2} = z^{2}$, then
$$g(y) = -\frac{y}{x} g(x) .$$
Letting $x = 4$ and $y = 3$, $(*)$ shows that $g(3) = 0$.
For any even number $x = 2n > 4$, let $y = n^{2} - 1$. Then $y > x$ and $x^{2} + y^{2} = \left(n^{2} + 1\right)^{2}$. For any odd number $x = 2n + 1 > 3$, let $y = 2(n + 1)n$. Then $y > x$ and $x^{2} + y^{2} = \left((n + 1)^{2} + n^{2}\right)^{2}$. Thus for every $x > 4$ there is $y > x$ such that $(*)$ is satisfied.
Suppose for a contradiction, that there is $x > 4$ with $g(x) > 0$. Then we can construct a
sequence $x = x_{0} < x_{1} < x_{2} < \ldots$ where $g\left(x_{i+1}\right) = -\frac{x_{i+1}}{x_{i}} g\left(x_{i}\right)$. It follows that $\left|g\left(x_{i+1}\right)\right| > \left|g\left(x_{i}\right)\right|$ and the signs of $g\left(x_{i}\right)$ alternate. Since $g(x)$ is always an integer, $\left|g\left(x_{i+1}\right)\right| \geq \left|g\left(x_{i}\right)\right| + 1$. Thus for some sufficiently large value of $i$, $g\left(x_{i}\right) < -f(0)$, a contradiction.
As for Proof 1, we now conclude that the functions that satisfy the given functional equation are $f(x) = c$, $c \in \mathbb{N}$.

Suppose that $W$ is the set of nonnegative integers and that $f: W \rightarrow W$ satisfies:
$$x f(y) + y f(x) = (x + y) f\left(x^{2} + y^{2}\right) .$$
We will show that $f$ is a constant function.
Let $f(0) = k$, and set $S = \{x \mid f(x) = k\}$.
Letting $y = 0$ in $(*)$ shows that $f\left(x^{2}\right) = k \quad \forall x > 0$, and so
$$x^{2} \in S \quad \forall x > 0$$
In particular, $1 \in S$.
Suppose $x^{2} + y^{2} = z^{2}$. Then $y f(x) + x f(y) = (x + y) f\left(z^{2}\right) = (x + y) k$. Thus,
$$x \in S \quad \text{iff} \quad y \in S$$
whenever $x^{2} + y^{2}$ is a perfect square.
For a contradiction, let $n$ be the smallest non-negative integer such that $f\left(2^{n}\right) \neq k$. By (l) $n$ must be odd, so $\frac{n-1}{2}$ is an integer. Now $\frac{n-1}{2} < n$ so $f\left(2^{\frac{n-1}{2}}\right) = k$. Letting $x = y = 2^{\frac{n-1}{2}}$ in $(*)$ shows $f\left(2^{n}\right) = k$, a contradiction. Thus every power of 2 is an element of $S$.
For each integer $n \geq 2$ define $p(n)$ to be the largest prime such that $p(n) \mid n$.
Claim: For any integer $n > 1$ that is not a power of 2, there exists a sequence of integers $x_{1}, x_{2}, \ldots, x_{r}$ such that the following conditions hold:
a) $x_{1} = n$.
b) $x_{i}^{2} + x_{i+1}^{2}$ is a perfect square for each $i = 1, 2, 3, \ldots, r-1$.
c) $p\left(x_{1}\right) \geq p\left(x_{2}\right) \geq \ldots \geq p\left(x_{r}\right) = 2$.
Proof: Since $n$ is not a power of $2$, $p(n) = p\left(x_{1}\right) \geq 3$. Let $p\left(x_{1}\right) = 2m + 1$, so $n = x_{1} = b(2m + 1)^{a}$, for some $a$ and $b$, where $p(b) < 2m + 1$.
Case 1: $a = 1$. Since $\left(2m + 1, 2m^{2} + 2m, 2m^{2} + 2m + 1\right)$ is a Pythagorean Triple, if $x_{2} = b\left(2m^{2} + 2m\right)$, then $x_{1}^{2} + x_{2}^{2} = b^{2}\left(2m^{2} + 2m + 1\right)^{2}$ is a perfect square. Furthermore, $x_{2} = 2b m(m + 1)$, and so $p\left(x_{2}\right) < 2m + 1 = p\left(x_{1}\right)$.
Case 2: $a > 1$. If $n = x_{1} = (2m + 1)^{a} \cdot b$, let $x_{2} = (2m + 1)^{a-1} \cdot b \cdot \left(2m^{2} + 2m\right)$, $x_{3} = (2m + 1)^{a-2} \cdot b \cdot \left(2m^{2} + 2m\right)^{2}$, $\ldots$, $x_{a+1} = (2m + 1)^{0} \cdot b \cdot \left(2m^{2} + 2m\right)^{a} = b \cdot 2^{a} m^{a}(m + 1)^{a}$. Note that for $1 \leq i \leq a$, $x_{i}^{2} + x_{i+1}^{2}$ is a perfect square and also note that $p\left(x_{a+1}\right) < 2m + 1 = p\left(x_{1}\right)$.
If $x_{a+1}$ is not a power of 2, we extend the sequence $x_{i}$ using the same procedure described above. We keep doing this until $p\left(x_{r}\right) = 2$, for some integer $r$.
By (2), $x_{i} \in S$ iff $x_{i+1} \in S$ for $i = 1, 2, 3, \ldots, r-1$. Thus, $n = x_{1} \in S$ iff $x_{r} \in S$. But $x_{r}$ is a power of 2 because $p\left(x_{r}\right) = 2$, and we earlier proved that powers of 2 are in $S$. Therefore, $n \in S$, proving the claim.
We have proven that every integer $n \geq 1$ is an element of $S$, and so we have proven that $f(n) = k = f(0)$, for each $n \geq 1$. Therefore, $f$ is constant, Q.E.D.