$ f \left( f ( x ) ^ 2 + y \right) = x ^2 + f ( y ) $ converts into Cauchy. 
Given the functional equation
$$ f \left( f ( x ) ^ 2 + y \right) = x ^ 2 + f ( y ) \text , $$
find all possible solutions.

I'm not exactly sure if it's continuous or bounded or what (since it's not mentioned). I've just assumed that but I would like if someone can prove or disprove it.
I've found that, assuming​ all the possible stuffs (as mentioned above), we get $ f ( 0 ) = 0 $, $ f ( a ) = f ( b ) \implies a ^ 2 = b ^ 2 $, $ f \left( f ( x ) ^ 2 \right) = x ^ 2 $ and using these, it reduces down to Cauchy's additive functional equation.
But I need to know if all those continuity or boundedness work here or not. If those aren't true here, then can't we apply Cauchy equation? When can we consider an equation to satisfy the Cauchy equation? And to use that, what do we require to prove before using Cauchy?
 A: For the sake of completeness, I will also repeat the arguments for the parts you've already achieved.
Let $ f : \mathbb R \to \mathbb R $ satisfy
$$ f \left( f ( x ) ^ 2 + y \right) = x ^ 2 + f ( y ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $. Setting $ y = 0 $ in \eqref{0} we get
$$ f \left( f ( x ) ^ 2 \right) = x ^ 2 + f ( 0 ) \text , \tag 1 \label 1 $$
which implies that if $ f ( x ) = f ( y ) $ then $ x ^ 2 = y ^ 2 $. In particular, if we put $ x = 0 $ in \eqref{1}, we get $ f \left( f ( 0 ) ^ 2 \right) = f ( 0 ) $, and thus $ f ( 0 ) ^ 4 = 0 ^ 2 $, or equivalently $ f ( 0 ) = 0 $. Using this together with \eqref{1} and substituting $ f ( x ) ^ 2 $ for $ x $ in \eqref{0}, we have
$$ f \left( x ^ 4 + y \right) = f ( x ) ^ 4 + f ( y ) \text . \tag 2 \label 2 $$
In particular, setting $ y = 0 $ in \eqref{2}, we get
$$ f \left( x ^ 4 \right) = f ( x ) ^ 4 \text . \tag 3 \label 3 $$
We can use \eqref{3} to rewrite the right-hand side of \eqref{2}, which will show that if $ x \ge 0 $, then $ f ( x + y ) = f ( x ) + f ( y ) $. Since for any $ x $ we have $ | x | \ge 0 $ and $ | x | + x \ge 0 $, thus we get
$$ f ( x + y ) = f \big( ( | x | + x ) + ( y - | x | ) \big) = f ( | x | + x ) + f ( y - | x | ) \\
= f ( | x | ) + f ( x ) + f ( y - | x | ) = \big( f ( | x | ) + f ( y - | x | ) \big) + f ( x ) \text , $$
and hence
$$ f ( x + y ) = f ( x ) + f ( y ) \text . \tag 4 \label 4 $$
Now, note that \eqref{3} implies $ f ( x ) \ge 0 $ when $ x \ge 0 $. This means that the function is increasing, as if $ x \le y $, we can substitute $ y - x $ for $ y $ in \eqref{4} to get $ f ( y ) = f ( x ) + f ( y - x ) \ge f ( x ) $. This, together with \eqref{4} implies that letting $ a = f ( 1 ) $, we have $ f ( x ) = a x $ for all $ x \in \mathbb R $. Plugging this into \eqref{1}, you'll find that the only solution is the identity function.
