A problem on implicit function theorem

Question

Does the function $x^2+y+\sin(xy)=0$ define $x$ implicitly as a differentiable function of $y$ locally around the point $(x,y)=(0,0)$?

Let $F(x,y)=0.$ Here $\frac{\partial F}{\partial x}|_{(0,0)}=0$. Thus can we say that it is impossible?

Answer 1

Yes, it's not possible.

Note though that this does not mean that you cannot solve $x =x (y)$, but it does mean you cannot do it in a differentiable way near $0$.

(To get an easier to grasp, one dimensional picture look at $y=x^3$. You can solve for $y$, but the result ist not differentiable in $0$)

(I do not say nor do I deny that in your example a (continuous) solution is possible)

Edit One additional note: the set $F=0$ is a smooth $1$-d submanifold ( a curve) of Euclidean two space in a neighbourhood of $0$, since $\frac{\partial F}{\partial y}(0,0)=1 \neq 0$ The two partial derivates in $0$ show this curve has a tangent pointing in direction of the $y$ axis, so if you try to write is as a graph over the $x$ axis that function will have infinite slope, so it cannot be differentiable.