Is $(x,y) \mapsto (\frac xy, x^2-y^3)$ an automorphism of $k(x,y)$? Let $k$ be a field of characteristic zero. The group of automorphisms of $k(x,y)$ is the cremona group; see also wikipedia.
If we are given a map $H: k(x,y) \to k(x,y)$, defined by 
$H(x):= f \in k(x,y)$ and $H(y):=g \in k(x,y)$ and extended to all $k(x,y)$ in a way that makes it a $k$-endomorphism of $k(x,y)$,
is there a criterion which tells if $H$ is an automorphism of $k(x,y)$? Probably in general it would be very difficult, so special cases are also of interest.
In particular, 

Is $H: (x,y) \mapsto (\frac xy, x^2-y^3)$ an automorphism of $k(x,y)$?

Define $\tilde{H}: k^2 \to k^2$ by $\tilde{H}(a,b) \mapsto (f(a,b),g(a,b))$.
Then for $t \in k^{\times}$: $\tilde{H}(t^3,t^2)=(\frac{t^3}{t^2}, (t^3)^2-(t^2)^3)=(t,0)$; perhaps this may help, I am not sure.
Any hints and comments are welcome!
 A: As long as the images of $x$ and $y$ are independent transcendentals in $k(x,y)$, your map is well-defined as a field morphism, and therefore one-to-one. But you need to check that the image is all of the codomain, i.e. that the map is onto. In other words, you need to express $x$ as a rational function of $x/y$ and $x^2-y^3$, similarly for $y$.
It’s not clear what “$f(t^2,t^3)$” can mean, till you tell me what element of $k(x,y)$ is equal to $t$.
EDIT- expansion:
It seems clear to me from your comments and from what I complained about above that you don’t fully apprehend the definition of your map $H$.
This is the way that $H$ is defined, using the given recipe $H(x)=x/y$, $H(y)=x^2-y^3$. It means that for a general rational function $\rho(x,y)$, described in the upper line of the display below, $H(\rho)$ is as described in the second line.
\begin{align}
\rho&=\frac{\sum_{i,\,j}a_{ij}x^iy\,^j}{\sum_{i,j}b_{ij}x^iy\,^j}\\
H(\rho)&=\frac{\sum_{i,\,j}a_{ij}(x/y)^i(x^2-y^3)\,^j}{\sum_{i,\,j}b_{ij}(x/y)^i(x^2-y^3)\,^j}
\end{align}
