Let $\mathbb{P}^{2}$ be a projective plane over some algebraically closed field. Suppose $C_{1}$ denote a curve defined by $y$ and $C_{2}$ be a curve defined by $y^{2}z-x^{3}$, where $[x:y:z]$ denotes a point in $\mathbb{P}^{2}$. How can I show that the bijective equivalence between $C_{1}$ and $C_{2}$?
I tried to solve the question by constructing a rational map $$f: C_{1} \to C_{2} \textrm{ by }[x:0:1] \mapsto [x^{2}: x^{3}:1] \textrm{ and } g: C_{2} \to C_{1} \textrm{ by } [x:y:1] \mapsto \left[\frac{y}{x}:0:1 \right].$$
And it works for one way; which is $g\circ f =1_{\mathbb{P}^{2}}$, an identity function. However, I failed to show that $f \circ g = 1_{\mathbb{P}^{2}}$. I think $$f \circ g ([a:b:1])= \left[\frac{a^{2}}{b^{2}}: \frac{a^{3}}{b^{3}}:1\right], $$ which is a little bit suspcious to say that it is identity.
Any help or hint would be appreciated.
In fact we have $$f\circ g([a:b:1])=\left[\frac{b^2}{a^2},\frac{b^3}{a^3},1\right]$$ On the domain of $f\circ g$ (i.e. on $C_2$) we have $a^3=b^2$. It follows that $b^2/a^2=a,b^3/a^3=b$.