Does $\frac{dx}{dy}$ at $0$ exist and $\frac{dy}{dx}\frac{dx}{dy}$ at $0$ is $1$ for the following function? $$y =f(x)=\begin{cases}x+x^2\sin\frac1{x^2}&\text{if }x\ne 0\\ 0&\text{if }x=0\end{cases}$$
We can see $\dfrac{dy}{dx}$ at $0$ is $1$. My question:

Does $\dfrac{dx}{dy}$ at $0$ exist and $\dfrac{dy}{dx}$ . $\dfrac{dx}{dy}$ at $0$ is $1$?

If not then  is it because $f$ can not be invertible in any nbd around the point $0$ ?
Actually  I was trying to understand when $\dfrac{dx}{dy}$ at a point exists and $\dfrac{dy}{dx}$ . $\dfrac{dx}{dy}$ at that point is $1$ if $\dfrac{dy}{dx}$ at that point exists. 
Can anyone please help me to clear  my doubt?
 A: Of course, for $dx/dy$ to exist at some point, we must have that $dy/dx>0$ or strictly negative at that point. However, this is not sufficient, since we need the function to be invertible in a neighborhood of this point; this happens if $dy/dx$ maintains its sign in a neighborhood of this point. Also, one can see that if $dy/dx$ does not maintain the same sign in a neighborhood of this point, then it cannot be strictly monotonic there.
Thus, what we need is that $f'$ have the same sign in a neighborhood of the point of concern.
To have this happen, it is sufficient that $f$ be continuously differentiable in a neighborhood of the point of interest; however, this is clearly too restrictive. Indeed, one sees that even if $f'$ is discontinuous at the point of concern but has a limiting value (possibly infinite), all we need require is that the limiting value have the same sign as the value of the derivative at the point in question. We may say more -- even if there is no limiting value, so long as the discontinuity is of the first kind (i.e., a jump), then again as long as the half-limits have the same sign as the derivative at the point in question, then all would still be well. Indeed, if either or both of the half-limits fails or fail to exist, then this is still doable provided that there is some neighborhood of the point where the derivative is of the same sign.
In these cases the derivative of the inverse is given by $$\frac{1}{dy/dx}.$$ To see the latter, since we have that $y=f(x)$ and $g=f^{-1}$ exists, then we may solve for $x$ to get $x=g(y).$ Now differentiating $y=f(x)$ with respect to $y$ gives $$dy/dy=\frac{d}{dx}f(x)\frac{dx}{dy},$$ or $$1=\frac{dy}{dx}\frac{dx}{dy},$$ and the result follows.
On the other hand, for some derivatives discontinuous at the point in question, they may change sign in every neighborhood of the point; then one can see that in such cases it is impossible to have an inverse mapping.
So, the function defined in OP, while having positive derivative at the origin, fails to have a continuous derivative near the origin, since for $x\ne 0,$ the derivative is given by $$1+2x\sin\frac{1}{x^2}-\frac2x\cos\frac{1}{x^2},$$ which has no limiting value at the origin. But we have seen that this is not perforce an issue. However, here, we see that the derivative oscillates between positive and negative values near the origin, and simultaneously becomes unbounded. Thus, it is not strictly monotonic in any interval about $x=0,$ and therefore cannot be invertible there. It follows that we cannot unambiguously calculate $dx/dy$ at that place.
