Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$:

$$
f\left(x^{2022}+y\right)=f\left(x^{1747}+2 y\right)+f\left(x^{42}\right)
$$
For complicated exponents, we will try to simplify them. For a given $x$, we have $x^{2022} + y = x^{1747} + 2y$ if and only if $y = x^{2022} - x^{1747}$, so we take this $y$. We then have $f\left(2 x^{2022} - x^{1747}\right) = f\left(2 x^{2022} - x^{1747}\right) + f\left(x^{42}\right)$, from which: $f\left(x^{42}\right) = 0$, thus $f$ is zero on the positives. Considering $y = 0$, we then have for all negative $x$, $f\left(x^{1747}\right) = 0$, hence $f = 0$. This function works $(0 + 0 = 0)$.