Differentiability at a point Let $f:\mathbb{R}^{2}\mapsto\mathbb{R}\mathbb{}^{2}$ be given
by 
$$f(x,y) = \left(\begin{array}{c}
x^{2}y+2y-x\\
3xy+4y
\end{array}\right)$$
Find a open set containing (0,0) where f has a differentiable inverse?.
I know the inverse function theorum guarentees there exists a neigbourhood
(open ball around) 
$(0,0)$ on which an inverse exists (since f has continuous partial derivatives) but in the sample solution for the example 
it has an additional point $\nabla f_{1}$ points in a direction in
the second quandrant and $\nabla f_{2}$ points in a direction in
the first quadrant, 
and so the level curves of $f_{1}$ and $f_{2}$ cannot cross twice
on this ball. 
Hence $f$ has an inverse on $B((0,0), 1/2)$- Why is is important that
the level curves do not cross (is it a theorum?) and how are they 
checking they $(f_1,f_2)$ point in different directions? .
 A: Let $U$ be an open  neighborhood of $(0,0)\in{\mathbb R}^2$ for which $d{\bf f}(x,y)$ is regular at all points $(x,y)\in U$. When ${\bf f}$ is injective on $U$ then ${\bf f}$ maps $U$ bijectively onto an open neighborhood $V$ of $(0,0)$, and the inverse map ${\bf f}^{-1}:\ V\to U$ is $C^1$, by the inverse function theorem.
The restriction ${\bf f}\restriction U$ is not injective iff there are two different points $(x_1,y_1)$,  $(x_2,y_2)\in U$, satisfying
$$f_1(x_1,y_1)=f_1(x_2,y_2)=:u_0,\quad f_2(x_1,y_1)=f_2(x_2,y_2)=:v_0\ .$$
But this means that the two level lines
$$f_1(x,y)=u_0,\qquad f_2(x,y)=v_0$$
intersect in two different points, both lying in $U$.
For a quantitative study of the ${\bf f}$ at hand we compose it with the linear map
$$\bigl(d{\bf f}(0,0)\bigr)^{-1}=\left[\matrix{-1&{\textstyle{1\over2}}\cr 0&{\textstyle{1\over4}}\cr}\right]$$
and obtain the new map
$${\bf g}:\quad (x,y)\mapsto\left\{\eqalign{u&=x+{\textstyle{3\over2}}xy-x^2y\cr v&=y+{\textstyle{3\over4}}xy\cr}\right.$$
with $$d{\bf g}(0,0)=\left[\matrix{1&0\cr 0&1\cr}\right]\ .$$
Compute
$$\nabla g_1(x,y)=\bigl(1+{\textstyle{3\over2}}y-2xy,{\textstyle{3\over2}}x-x^2\bigr),\qquad \nabla g_2(x,y)=\bigl({\textstyle{3\over4}}y, 1+{\textstyle{3\over4}}x\bigr)\ .$$
When $|x|, \>|y|<{1\over20}$ (this bound should do the job), then the vector $\nabla g_1$ includes an angle $<45^\circ$ with the $x$-axis. This implies that the level lines of $g_1$ (resp., their tangents) include an angle $<45^\circ$ with the $y$-axis. In a similar way the level lines of $g_2$ include an angle $<45^\circ$ with the $x$-axis. It should then be clear that (a) the two gradients are always linearly independent, and  that (b) a $g_1$- and a $g_2$-level-line can intersect in at most one point. It follows that ${\bf f}$ is injective in the open square $\>\bigl]-{1\over20},{1\over20}\bigr[\times\bigl]-{1\over20},{1\over20}\bigr[\>$.
By the way: The equations $f_1(x,y)=u$, $\>f_2(x,y)=v$ can be solved explictly for $(x,y)$ in the neighborhood of $(x,y)=(u,v)=(0,0)$ in terms of square roots.
