Functional equation: $ f \left( x + y ^ 2 + z ^ 3 \right) = f ( x ) + f ^ 2 ( y ) + f ^ 3 ( z ) $ 
Exercise: Find all of functions $f: \mathbb{R} \to \mathbb{R}$ such that $$ f \left( x + y ^ 2 + z ^ 3 \right) = f ( x ) + f ^ 2 ( y ) + f ^ 3 ( z ) \text , \forall x , y , z \in \mathbb R \text . $$

In fact, I solved this question as follows:

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*First step: Calculate $ f ( 0 ) $. In this step, I obtained that $ f ( 0 ) = 0 $ or $ f ( 0 ) = - 1 $.

*Second step: Consider two cases of the first step. I determinated the solutions are $ f ( x ) = 0 $, $ f ( x ) = - 1 $ and $ f ( x ) = x $.

The problem is that this way is very long. Who has another way to solve my exercise?
 A: Well this is not the most elegant of proofs either...
Case where $f(0) = 0$
We see that $f(x + y) = f(x + 0^2 + (\sqrt[3]y)^3) = f(x) + f(0)^2 + f(\sqrt[3]y)^3 = f(x) + f(y)$.
We also see that if $x \leq y$ then $f(x) \leq f(y)$ because $f(y) = f(x + y - x) = f(x + \sqrt{y - x}^2 + 0^3) = f(x) + f(\sqrt{y - x})^2 + f(0)^3 \geq f(x)$.
From the first relation above, we find that $f\left(\frac{1}{n}\right) = \frac{f(1)}{n}$ for $n \in \mathbb{N}^*$.
For any $\epsilon \gt 0$, choose $n$ large enough so that $\frac{|f(1)|}{n} \lt \epsilon$. Choose $\delta = \frac{1}{n}$.
Then whenever $|x| \lt \delta$, we find that $|f(x)| \lt |f(\delta)| = |f(1/n)| = \frac{|f(1)|}{n} \lt \epsilon$ which shows that $\lim\limits_{x \rightarrow 0} f(x) = f(0) = 0$.
Finally we see that $\lim\limits_{h \rightarrow 0}f(x + h) - f(x) = \lim\limits_{h \rightarrow 0}f(h) = 0$ so $f$ is continuous everywhere.
$\frac{f(x)}{x}=\frac{f(rx)}{rx}$ for all $r \in \mathbb{Q}$ shows that $f(x)/x$ is constant on a dense subset of $\mathbb{R^+}$, and by continuity and even-ness, must be constant everywhere on $\mathbb{R}^*$.
So on $\mathbb{R}^*$ we have $f(x) = Ax$ for some $A$ which can be found to satisfy $-A = A(-1 + (-1)^2 + (-1)^3) = -A + A^2 - A^3$ or $A \in \{0, 1\}$.
So $f(x) = x$ or $f(x) = 0$ on $\mathbb{R}^*$, and thus also on all of $\mathbb{R}$.

Not yet worked out the case where $f(0) = -1$.
A: The case $ f ( 0 ) = 0 $ is solved by @TobEnrack. For the case $ f ( 0 ) = -1 $, let $ x = y = 0 $ in
$$ f \big( x + y ^ 2 + z ^ 3 \big) = f ( x ) + f ( y ) ^ 2 + f ( z ) ^ 3 \tag 0 \label 0 $$
and you get
$$ f \big( z ^ 3 \big) = f ( z ) ^ 3 \tag 1 \label 1 $$
Again, letting $ x = z = 0 $ in \eqref{0} you have
$$ f \big( y ^ 2 \big) = f ( y ) ^ 2 - 2 \tag 2 \label 2 $$
Now combining \eqref{1} and \eqref{2} you get
$$ f ( x ) ^ 6 - 2 = \big( f ( x ) ^ 3 \big) ^ 2 - 2 = f \big( x ^ 3 \big) ^ 2 - 2 = f \Big( \big( x ^ 3 \big) ^ 2 \Big) = f \big( x ^ 6 \big) \\ = f \Big( \big( x ^ 2 \big) ^ 3 \Big) = f \big( x ^ 2 \big) ^ 3 = \big( f ( x ) ^ 2 - 2 \big) ^ 3 = f ( x ) ^ 6 - 6 f ( x ) ^ 4 + 12 f ( x ) ^ 2 - 8 $$
By simple algebra, it's easy to derive
$$ f ( x ) ^ 2 = 1 \tag 3 \label 3 $$
Finally, letting $ z = 0 $ in \eqref{0} and using \eqref{3} you get
$$ f \big( x + y ^ 2 \big) = f ( x ) \tag 4 \label 4 $$
Hence, if $ x \le y $, substituting $ \sqrt { y - x } $ for y in \eqref{4}, it's easy to see that
$$ f ( y ) = f \big( x +  ( y - x ) \big) = f \Big( x + { \sqrt { y - x } } ^ 2 \Big) = f ( x ) $$
Since for every $ x $, either $ x \le 0 $ or $ 0 \le x $, therefore $ f ( x ) = -1 $.
