Let $\mathbb P^2$ be the Projective plane over an algebraically closed field $k$.
Is the rational map $f :\mathbb P^2 \to \mathbb P^2$ given by $f(x: y : z)=(x^2 : y^2: z^2) $ Birational ?
In general, if $f(x: y: z)=(f_0(x,y,z) : f_1(x,y,z) : f_2(x,y,z) )$ is a rational map on $\mathbb P^2$ where $f_i$ 's are non-zero monomials of same degree with no common factor, then when can we say that $f$ is Birational ? What happens if all the $f_i$ are also square-free ?
The map in your example is regular and finite of degree 4, hence not birational.
If every $f_i$ is a monomial of degree $d_i = (d_{i0},d_{i1},d_{i2})$, the map is toric, in paricular it restricts to a group endomorphism of the algebraic torus $$ T = \{ x \ne 0, y \ne 0, z\ne 0\} \subset \mathbb{P}^2. $$ In particular, it is birational if and only if the endomorphism is an isomorphism, i.e., if and only if $$ \det\begin{pmatrix} d_{00} & d_{01} & d_{02} \\ d_{10} & d_{11} & d_{12} \\ d_{20} & d_{21} & d_{22} \end{pmatrix} = \pm d. $$