Find all the differentiable functions $f$ with the property $f\left(x^4+y\right) = x^3f(x) + f(y)$ I'm having a hard time answering this question:

Find all the differentiable functions $f$ such that
$$\forall x,y\in\mathbb R : f\left(x^4+y\right) = x^3f(x) + f(y) \text.$$

I would love to get a hint about how to start since this is my first time answering this type of questions.
Thanks a lot.

You can show that a function $ f : \mathbb R \to \mathbb R $ satisfies
$$ f \left( x ^ 4 + y \right) = x ^ 3 f ( x ) + f ( y ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ iff there is a constant $ a \in \mathbb R $ such that $ f ( x ) = a x $ for all $ x \in \mathbb R $, even without the assumption of differentiability. It's easy to check that functions of this form are solutions. We try to prove the converse.
Put $ y = 0 $ in \eqref{0} to get $ f \left( x ^ 4 \right) = x ^ 3 f ( x ) + f ( 0 ) $, which in particular for $ x = 1 $ shows that $ f ( 0 ) = 0 $ and hence
$$ f \left( x ^ 4 \right) = x ^ 3 f ( x ) \text . \tag 1 \label 1 $$
This lets you rewrite \eqref{0} as
$$ f \left( x ^ 4 + y \right) = f \left( x ^ 4 \right) + f ( y ) \text . $$
This means that for all $ x \in \mathbb R ^ { 0 + } $ and all $ y \in \mathbb R $ we have $ f ( x + y ) = f ( x ) + f ( y ) $. Noting that for any $ x \in \mathbb R $ we have $ | x | \ge 0 $ and $ x + | x | \ge 0 $, we can generalize this and get
$$ f ( x + y ) = f \big( ( | x | + x ) + ( y - | x | ) \big) = f ( | x | + x ) + f ( y - | x | ) \\
= f ( | x | ) + f ( x ) + f ( y - | x | ) = f ( x ) + f \big( | x | + ( y - | x | ) \big) \text , $$
and thus
$$ f ( x + y ) = f ( x ) + f ( y ) \tag 2 \label 2 $$
for all $ x , y \in \mathbb R $. Letting $ y = - x $ in \eqref{2} we have
$$ f ( - x ) = - f ( x ) \tag 3 \label 3 $$
and using induction and \eqref{2} we get
$$ f ( n x ) = n f ( x ) \tag 4 \label 4 $$
for every positive integer $ n $. Defining $ a = f ( 1 ) $, we can use \eqref{1} and \eqref{2} to get
$$ f \left( ( x + 1 ) ^ 4 \right) = ( x + 1 ) ^ 3 f ( x + 1 ) = ( x + 1 ) ^ 3 \big( f ( x ) + a \big) \text . \tag 5 \label 5 $$
Again, by \eqref{1}, \eqref{2} and \eqref{4} we have
$$ f \left( ( x + 1 ) ^ 4 \right) = f \left( x ^ 4 + 4 x ^ 3 + 6 x ^ 2 + 4 x + 1 \right) \\
= f \left( x ^ 4 \right) + f \left( 4 x ^ 3 \right) + f \left( 6 x ^ 2 \right) + f ( 4 x ) + a \\
= x ^ 3 f ( x ) + 4 f \left( x ^ 3 \right) + 6 f \left( x ^ 2 \right) + 4 f ( x ) + a \text . \tag 6 \label 6 $$
Combining \eqref{5} and \eqref{6} we get
$$ \left( x ^ 3 + 3 x ^ 2 + 3 x + 1 \right) \big( f ( x ) + a \big) = x ^ 3 f ( x ) + 4 f \left( x ^ 3 \right) + 6 f \left( x ^ 2 \right) + 4 f ( x ) + a \text . \tag 7 \label 7 $$
Similarly, calculating $ f ( x - 1 ) $ in two different ways, and this time also using \eqref{3}, we get
$$ \left( x ^ 3 - 3 x ^ 2 + 3 x - 1 \right) \big( f ( x ) - a \big) = x ^ 3 f ( x ) - 4 f \left( x ^ 3 \right) + 6 f \left( x ^ 2 \right) - 4 f ( x ) + a \text , $$
which together with \eqref{7} yields
$$ \left( x ^ 3 + 3 x \right) f ( x ) + a \left( 3 x ^ 2 + 1 \right) = x ^ 3 f ( x ) + 6 f \left( x ^ 2 \right) + a \text , $$
or simply
$$ f \left( x ^ 2 \right) = \frac 1 2 x f ( x ) + \frac a 2 x ^ 2 \text . $$
Similar to before, we calculate $ f \left( ( x + 1 ) ^ 2 \right) $ in two ways:
$$ f \left( ( x + 1 ) ^ 2 \right) = \frac 1 2 ( x + 1 ) \big( f ( x ) + a \big) + \frac a 2 ( x + 1 ) ^ 2 \text ; $$
$$ f \left( ( x + 1 ) ^ 2 \right) = f \left( x ^ 2 \right) + f ( 2 x ) + a = \frac 1 2 x f ( x ) + \frac a 2 x ^ 2 + 2 f ( x ) + a \text ; $$
and these together simplify to $ f ( x ) = a x $.
