Answer in 2 steps?

Putting $x = y = 0$ , we get $f(0) = 0$ .

Putting $y = 0$ , we have $f(x^2) = x^4$ . Putting $z = x^2$ , we have $f(z) = z^2$ for all $z \geq 0$ .

Therefore, $f(2015) = 2015^2 = (5 * 13 * 31)^2.$

Therefore answer $= \boxed{51}$

$1^{st}$ step-Put $x=y=0$ and we get $f(0)=0$

$2^{nd}$ step-Put $x=\sqrt{2015},y=0$ and we get $f(2015)=31^{2}\times13^{2}\times5^{2}$

13+31+5+2=51

A 2 step solution.

The function is simply (x^2)

so answer is 2015^2

and 2015= (13)(31)(5)

squaring it we have answer as (a,b,c,d)=(13,31,5,2)

The problem with functional equations is that they don't fit this form of problems with a numerical short answer...

Stupid solution: Observe that $f(x) = x^2$ satisfies the functional equation. Since this problem exists, we assume that the answer is the same no matter what $f$ we choose. Thus $f(2015) = 2015^2 = 5^2 13^2 31^2$ , giving $a+b+c+d = 5+13+31+2 = \boxed{51}$ .

Actual solution is given by Samuraiwarm Tsunayoshi.

The answer doesn't change if we use real numbers; $f(x^2) = x^2f(0)+x^4$ means $f(x) = xf(0)+x^2$ for all $x \ge 0$ , and since $2015 \ge 0$ , the result remains. But the function solution itself doesn't change; if we plug in $x \gets -x$ and compare it with the original, we get that everything cancels out except $y^2 f(x) = y^2 f(-x)$ . Taking $y=1$ , we get $f(x) = f(-x)$ , so $f$ can be extended to the negative reals (which is still $f(x)=x^2$ ).

Plugging $y = 0$ we get

$f(x^{2}) = x^{2}f(0) + x^{4}$ (1)

$f(x) = xf(0) + x^{2}$ (2)

Plugging into the original equation we get

$(x^{2}+y^{2})f(0) + (x^{2}+y^{2})^{2} = y^{2}(xf(0)+x^{2}) + x^{2}(yf(0) + y^{2}) + x^{4} + y^{4}$

$2x^{2}y^{2} + x^{2}f(0) + y^{2}f(0) = xy(x+y)f(0) + 2x^{2}y^{2}$

$(x^{2}+y^{2} - xy(x+y))f(0) = 0$

$f(0) = 0$

Therefore, $f(x) = \boxed{x^{2}}$ . ~~~