Solve functional equation $f(x^4+y)=x^3f(x)+f(y)$ I need help solving this equation, please: 
$$f(x^4+y)=x^3f(x)+f(y),$$ such that $ f:\Bbb{R}\rightarrow \Bbb{R},$ and differentiable
I've found that $f(0)=0$ and $f(y+1)=f(1)+f(y)$, but I couldn't continue, I think the solution is $f(x)=ax$. 
Thanks for your help

To help clarify or expand the hint by @xidgel and avoid any possible confusion.
First introduce $t(x,y) = x^4+y$, then we can write:
$$f(t) = x^3f(x)+f(y)$$
Now we can express differentiation with respect to x and y with the chain rule:
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial t} \cdot \frac{\partial t}{\partial x} \text{ and } \frac{\partial f}{\partial y} = \frac{\partial f}{\partial t} \cdot \frac{\partial t}{\partial y}$$
Now proceed as in the hint by calculating these differentials on both sides of the equation above.
