Prove that $f(x,y)=(3x+x^3\exp(y),y-x^2)$ is a diffeomorphism $f:\mathbb R^2 \to \mathbb R^2$
$$f(x,y)=(3x+x^3\exp(y),y-x^2)$$
How can I show that $f$ is a diffeomorphism? The following hint was given: To show that $f$ is bijective one should show that $(u,v)=(3x+x^3\exp(y),y-x^2)$ has an unique solution.
 A: By definition, diffeomorphism is a bijective differentiable map, which inverse is also differentiable. The Inverse Function Theorem states that if a Jacobian (matrix of partial derivatives) of a differentiable map has a non-zero determinant, than there exists an inverse map, which is also differentiable. So it suffices to show that such defined $f(x,y)$ is one-to-one (injective), onto (surjective) and $\mbox{det}(Jf)\neq0$.
Let $f(3x+x^3e^y,y-x^2)=(u,v)$ for arbitrary $(u,v)\in\mathbb{R}^2$, that is equivalent to the system of equations
$$\begin{cases}3x+x^3e^y=u\\y-x^2=v\end{cases}\Longleftrightarrow\begin{cases}3x+x^3e^{v+x^2}=u\\y=v+x^2\end{cases}$$
The LHS of the equation $3x+x^3e^{v+x^2}=u$ is a monotonically increasing continuous function, therefore this equation has a unique solution for any given $u$, and for unique $x$ and any given $v$ we have uniqueness of $y$ from $y=v+x^2$. Thus $f$ is one-to-one and onto, hence bijective.
Let $Jf$ be the matrix of partial derivatives of $f$, then
$$\mbox{det}(Jf) = \mbox{det}\begin{pmatrix}\frac{\partial f_1}{\partial x}&&\frac{\partial f_1}{\partial y}\\ \frac{\partial f_2}{\partial x}&&\frac{\partial f_2}{\partial y}\end{pmatrix} = \mbox{det}\begin{pmatrix}3+3x^2e^y&&x^3e^y\\-2x&&1\end{pmatrix} = 3 + \underbrace{3x^2e^y + 2x^4e^y}_{\geq 0}\geq3\neq0.$$
Thus it follows from the Inverse Function Theorem that $f$ is a diffeomorphism.
