Cauchy-Riemann Equations + Locally Invertible implies non-zero derivative Suppose $C^1$ $f:\mathbb{R}^2 \to \mathbb{R}^2$ satisfies the Cuachy-Riemann equations $\frac{\partial f_1}{\partial x} = \frac{\partial f_2}{\partial y}, \frac{\partial f_1}{\partial y} = -\frac{\partial f_2}{\partial x}$ at some point $a \in \mathbb{R}^2$. Additionally, let $f$ be locally invertible. Is it true that $Df(a) \neq 0$? I have been asked to prove that in a homework, but I cannot figure out how to do it without further conditions (e.g. $f^{-1}$ is differentiable). Any tips would be greatly appreciated.

It seems like there are obvious counter examples as given, like $f(x,y) = (x^3, y^3)$.
The function $$f:\quad{\mathbb R}^2\to{\mathbb R}^2,\qquad(x,y)\mapsto(x^3,y^3)$$
is $C^1$, and even globally invertible, if we define (as usual) $$\root 3\of t:={\rm sgn}(t)\root3\of{|t|}\qquad(t\in{\mathbb R})\ .$$
Furthermore we have $$f_{1.1}=f_{2.2}=0,\qquad f_{1.2}=-f_{2.1}=0$$
at $(0,0)$, and therefore $Df(0,0)=0$.

The claim in your homework (as it appears in the question) can therefore not be proven.
I think the professor in our Honors Analysis class stated that local invertibility assumes a smooth inverse which (x^3,y^3) does not have at (0,0).