I am trying to prove that the map $\phi:P^1\to X = Z(x^2y^3-z^5)$, given by $[r:s]\mapsto [u^5:v^5:u^2v^3]$ is a birational equivalence, i.e. that there exists some map $\psi:X\to P^1$ such that $(\phi\circ \psi)[x:y:z]=[x:y:z]$ and $(\psi\circ \phi)[u:v]=[u:v]$.
As far as I understand, I have to get some rational expressions in $x=u^5$, $y=v^5$, and $z=u^2v^3$ that simplify to just $u$ and $v$. However, no matter how I try to multiply and add these expressions, I can't get $u$ and $v$ out of them. The simplest expressions I can get are $\frac uv=\frac{xy}{z^2}$ and $\frac vu=\frac{z^2}{xy}$, but that isn't exactly what I want. How do I get $u$ and $v$ by only using rational functions?
You are already almost there! Remember that the maps need not be defined everywhere, and $[u:v] = [u/v:1]\in\mathbb P^1$.
Let $U\subseteq X$ be the open set where $z\neq 0$, and $V\subseteq\mathbb P^1$ be the open set where $v\neq 0$. The map $\varphi:U\to V, [x:y:z]\to[xy/z^2:1] = [xy:z^2]$ is well-defined and regular on $U$. We can show that $\varphi$ is a birational inverse to $\phi$ by computing the composition explicitly on these open sets.
$\phi\circ\varphi([x:y:z]) = \phi([xy:z^2]) = [(xy)^5:(z^2)^5:(xy)^2(z^2)^3].$ The latter equals $[x^3y^2z^5:x^2y^3z^5:x^2y^2z^6]$ since $x^2y^3 = z^5$, which gives exactly $[x:y:z]$ after factoring $x^2y^2z^5$ out. Thus $\phi\circ\varphi = \operatorname{id}_U$.
Checking the other composition is trivial.