Existence of smooth function with prescribed zeros and value of derivative at the origin Let
$$
f(x) = \begin{cases}
e^{-1/x} & x > 0
\\
0 & x \leq 0.
\end{cases}
$$
Does there exist an infinitely differentiable function $h: \mathbb{R}^2 \to \mathbb{R}$ such that
$$
h(x, 0) = 0, \quad h(x, f(x)) = 0, \quad x \in \mathbb{R}
$$
and ${\partial h \over \partial y}(0, 0) \neq 0$?
I do not seem to be able to come up with an example.
My attempts at the problem

*

*Taking derivatives we get the following:
$$
{\partial h \over \partial x}(x, 0) = 0 \quad (x \in \mathbb{R}), \quad  {\partial h \over \partial x}(x, e^{-1 \over x}) + e^{-1 \over x}{1 \over x^2}{\partial h \over \partial y}(x, e^{-1 \over x}) = 0  \quad (x > 0),
$$
however neither equation gives information about the value of the partial derivatives of $h$ with respect to y at the origin.


*The function $h(x, y) = y(y - f(x))$ and other examples I have tried do not work.


*Using Taylor's theorem,
$$
h(x, y) = \partial_yh(0, 0)y  + o(\|(x, y)\|),
$$
and solving $y = e^{-1/x}$ for $x$, we get
$$
0 = h\left ({-1 \over \ln y}, y\right) = \partial_yh(0, 0)y + o\left(\sqrt{y^2 + {1 \over (\ln y)^2}}\right).
$$
The equation remains valid if $\partial_yh(0, 0) \neq 0$ because it is true that
$$
y = o\left(\sqrt{y^2 + {1 \over (\ln y)^2}}\right), \quad y \to 0.
$$
This doesn't lead to a contradiction.


*I've tried replacing $f(x)$ with other functions which vanish to order two at zero, but I get nothing better.

 A: Assume $h$ is $C^1$ and satisfies $h(x, 0) = 0$, and $h(x, f(x)) = 0$ for any $x\in\Bbb R$.
Since $h(x, 0) = 0$, and $h(x, f(x)) = 0$ for any $x >0$, there is $0<y_x<f(x)$ such that $\frac{\partial f}{\partial y}(x,y_x)=0$ (Rolle's theorem). Now, if $x\to 0+$ we get $(x,y_x)\to 0$, so
$$
\frac{\partial h}{\partial y}(0,0)\leftarrow\frac{\partial h}{\partial y}(x,y_x)=0,
$$
so $\frac{\partial h}{\partial y}(0,0)=0$.
A: First let's write down the definition of the partial derivative of $h$ wrt $y$.
$$\require{\cancel}
\frac{\partial h}{\partial y}=\lim_{j\to 0}\frac{h(x,y+j)-h(x,y)}{j}$$
Now, evaluate that at the point $(0,0)$.
$$\left.\frac{\partial h}{\partial y}\right|_{(0,0)}=\lim_{j\to 0}\frac{h(0,j)-\cancelto{0}{h(0,0)}}{j}$$
Notice $h(0,0)=0$ by $h(x,0)=0\Longleftarrow\left.h(x,0)\right|_{x=0}=h(0,0)$.
$$=\lim_{j\to 0}\frac{h(0,j)}{j}\neq 0$$
Now, consider the left-side limit:
$$\lim_{j\to 0^-}\frac{h(0,j)}{j}\neq 0$$
Notice that $\lim_{x\to0^-}f(x)=0$, so it's valid to substitute $f(x)$ for $j$.
$$\lim_{x\to 0^-}\frac{\cancelto{0}{h(x,f(x))}}{x}\neq 0$$
And thus, the statement is false.
