Functional equation problem: $ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) $ This functional equation problem is from the Latvian Baltic Way team selection competition 2019:

Find all functions $ f : \mathbb R \to \mathbb R $ such that for all real $ x $ and $ y $,
$$ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 +  f \left( x ^ 2 y + y \right) \text . \tag 1 \label {eqn1} $$

OK, so I think that the only answer is $ f ( x ) = 0 $.
I just want to see if my proof that it is the only solution is correct.
So we start off by plugging $ y = - y $. We get that
$$ f \left( y ^ 2 - f ( x ) \right) = - y f ( x ) ^ 2 + f \Big( - y \left( x ^ 2 + 1 \right) \Big) \text . $$
Then we add the two equations together getting that
$$ 2 f \left( y ^ 2 - f ( x ) \right) = f \Big( - y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) \text . $$
From the above equations we get that
$$ \frac { f \Big( - y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) } 2 = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \text. \tag 2 \label {eqn2} $$
Now, if we plug $ x = - x $ then we will get that the LHS is the same and that the RHS is $ y f ( - x ) ^ 2 + f \left( x ^ 2 y + y \right) $.
So we proceed by subtracting the two and getting that
$$ 0 = y f ( x ) ^ 2 - y f ( - x ) ^ 2 \text . $$
So, lets assume that $ y \ne 0 $ getting that
$$ 0 = \big( f ( x ) - f ( - x ) \big) \big( f ( x ) + f ( - x ) \big) \text . $$
Now we do a two case analysis, 1) the function is even and 2) the function is odd.
Lets start with the function being even then from \eqref{eqn2} we get that
$$ 0 = y f ( x ) ^ 2 \text , $$
which of course implies that the function is just $ 0 $.
OK, now the odd case. Since the function is odd, $ f ( 0 ) = 0 $. Then plugging $ x = 0 $ in \eqref{eqn1} we get that
$$ f \left( y ^ 2 \right) = f ( y ) \text , $$
which implies that $ f $ is also an even function. Since $ f $ is both even and odd, it can only be $ 0 $.
Since we got that $ f $ is zero in both cases, the only solution to the equation is $ f ( x ) = 0 $.

You can show that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is the constant zero function, by continuing your own argument. You've established showing
$$ f ( - x ) ^ 2 = f ( x ) ^ 2 \text . \tag 1 \label 1 $$
Now, take the "$ y f ( x ) ^ 2 $" term in \eqref{0} to the left-hand side, square both sides and rearrange the terms to get
$$ f \left( y ^ 2 - f ( x ) \right) ^ 2 + y ^ 2 f ( x ) ^ 4 - f \left( x ^ 2 y + y \right) ^ 2 = 2 y f ( x ) ^ 2 f \left( y ^ 2 - f ( x ) \right) \text . \tag 2 \label 2 $$
Substituting $ - y $ for $ y $ in \eqref{2} and using \eqref{1}, the left-hand side will be equal to the left-hand side of \eqref{2}, while the right-had side will be the opposite of the right-hand side of \eqref{2}. This in particular implies that
$$ y f ( x ) ^ 2 f \left( y ^ 2 - f ( x ) \right) = 0 \text . \tag 3 \label 3 $$
Now, if for some $ a \in \mathbb R $ we have $ f ( a ) \ne 0 $, then setting $ x = a $ and considering all nonzero values for $ y $ in \eqref{3}, we can see that for all $ z > - f ( a ) $ we have $ f ( z ) = 0 $. But then letting $ x = a $ and $ y = 1 + \frac { | f ( a ) | } { 1 + a ^ 2 } $ in \eqref{0} leads to a contradiction with $ f ( a ) \ne 0 $, since we have $ y > 0 $, $ y ^ 2 - f ( a ) > - f ( a ) $ and $ a ^ 2 y + y > - f ( a ) $. Therefore $ f ( x ) $ must be equal to $ 0 $ for all $ x \in \mathbb R $.
