Does $f(x,y)=(x^2y+x,6x+y^2)$ have a local inverse at $(1,1)$? Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be given by 
$$
f(x,y)=(x^2y+x,6x+y^2).
$$
A direct calculation shows that $\det Df(1,1)=0$. The key assumption in the Inverse Function Theorem is not satisfied. 
Here is my question:

Does there exist a local inverse of $f$ at $(1,1)$?

 A: Since $f(x,y)=(x^2y+x,6x+y^2)$, we have
$$
\frac{\partial\,f}{\partial(x,y)}
=\begin{bmatrix}2xy+1&6\\x^2&2y\end{bmatrix}
$$
and
$$
\det\begin{bmatrix}2xy+1&6\\x^2&2y\end{bmatrix}=4xy^2+2y-6x^2
$$
we get
$$
\det\frac{\partial\,f(1,1)}{\partial(x,y)}=0
$$
Thus, the Jacobian is singular at $(1,1)$.
Since
$$
\begin{bmatrix}1&-3\end{bmatrix}\begin{bmatrix}3&6\\1&2\end{bmatrix}=\begin{bmatrix}0&0\end{bmatrix}
$$
the singular direction is $\begin{bmatrix}1&-3\end{bmatrix}$. Looking at
$$
f(1+t,1-3t)=\left(2-5t^2-3t^3,7+9t^2\right)
$$
we want to subtract $(2,7)$ and rotate with $\begin{bmatrix}-5&9\\9&5\end{bmatrix}$ to align the cusp with the $x$-axis.
$$
(f(1+t,1-3t)-(2,7))\begin{bmatrix}-5&9\\9&5\end{bmatrix}=\left(t^2(106+15t),-27t^3\right)
$$
Adjusting the direction a very small amount, we see that
$$
(f(1+t,1-(3-a)t)-(2,7))\begin{bmatrix}-5&9\\9&5\end{bmatrix}\approx(106t^2,-27t^3+19at)
$$
which loops and intersects itself at $t\approx\pm\sqrt{\frac{19a}{27}}$.
Here is the curve for $a=0.1$:

The point of self-intersection is near the approximated point $(7.45926,0)$
Thus, no matter how close we get to $(1,1)$, there is the image of a line that gets mapped to a loop which intersects itself. Therefore, the function is not invertible near $(1,1)$.
