I'm struggling with the following problem:
Let $f\colon\mathbb{R}^2\to\mathbb{R}$ be a twice continuously differentiable function satisfying $$f(0,y)=0\mbox{ for all }y\in\mathbb{R}$$
(a) Show that $f(x,y) = xg(x,y)$ for all pairs $(x,y)\in\mathbb{R}^2$, where $g$ is the function given by $$g(x,y) = \int\limits_0^1\frac{\partial f}{\partial x}(tx,y)dt$$
(b) Show that $g$ is continuously differentiable and that for all $x\in\mathbb{R}$ $$g(0,y) = \frac{\partial f}{\partial x}(0,y),~\frac{\partial g}{\partial y}(0,y) = \frac{\partial^2 f}{\partial x\partial y}(0,y)$$
(c) If $\frac{\partial f}{\partial x}(0,0)\neq0$ there is a neighborhood $V$ of $(0,0)$ in $\mathbb{R}^2$ such that $f^{-1}(0)\cap V = V\cap \{x=0\}$
(d) If $\frac{\partial f}{\partial x}(0,0) = 0$ and $\frac{\partial^2 f}{\partial x\partial y}(0,0)\neq0$ there is a neighborhood $V$ of $(0,0)$ in $\mathbb{R}^2$ such that $f^{-1}(0)\cap V$ consists of the union of the set $V\cap\{x=0\}$ with a curve through $(0,0)$, whose tangent at $(0,0)$ is not vertical (not parallel to the $y$-axis)
Here are my attempts:
(a) Note that $\frac{\partial f}{\partial x}(tx,y) = \frac{t}{x}\cdot\frac{\partial f}{\partial t}(tx,y)$ and so using the integration by parts $$g(x,y) = \int\limits_0^1\frac{\partial f}{\partial x}(tx,y)dt = \int\limits_0^1 \frac{t}{x}\cdot\frac{\partial f}{\partial t}(tx,y)dt = \frac{1}{x}\left(f(x,y) - \int\limits_0^1 f(tx,y)dt\right)$$ it suffices to show that $\int\limits_0^1 f(tx,y)dt = 0$, however, I can't see how to proceed with it.
(b) Expressions for $g(0,y)$ and $\frac{\partial g}{\partial y}(0,y)$ is the result of a straightforward calculation using the definiton of $g(x,y)$, but I have troubles showing that $g$ is continuously differentiable. What am I supposed to do -- prove that the partial derivatives exist and they are continuous? If so, can someone write down the details?
(c) One can fix $y=0$ and consider a function $f(x,0)\colon\mathbb{R}\to\mathbb{R}$. Then, since $\frac{\partial f}{\partial x}(0,0)\neq0$, the Inverse Function Theorem says that there exists a neighborhood $U$ of $0\in\mathbb{R}$ such that $f\colon U\to f(U)$ is the bijection. Denote $V = U\times\{0\}$, so $V\subset\mathbb{R}^2$ and we are done. Am I right here?
(d) It seems like using the result of part (b) we can do the same trick with the function $g(0,y)\colon\mathbb{R}\to\mathbb{R}$, but I can't finish the proof.
Any help will be really appreciated. Thanks a lot in advance.
I think you are misunderstanding something with the integral. $\frac{\partial f}{\partial x}(tx,y)$ doesn't mean the derivative with respect to the $x$ inside, it means the derivative of the function with respect to the first argument. In other words the integral really evaluates to
$$g(x,y) = \int_0^1 \frac{\partial f}{\partial x}(tx,y)\:dt = \begin{cases} \frac{1}{x}f(tx,y)\Bigr|_0^1 = \frac{f(x,y)-f(0,y)}{x} = \frac{1}{x}f(x,y) & x\neq0 \\ \int_0^1 \frac{\partial f}{\partial x}(0,y) \:dt = \frac{\partial f}{\partial x}(0,y) & x= 0 \\ \end{cases}$$
therefore we have in both cases that $f(x,y) = xg(x,y)$.
In order to show $g$ is continuous differentiable it is enough (sufficient but not necessary) to show that the partial derivatives of $g$ exist and are continuous.
The other questions are just application of the inverse function theorem on $g$ and its first partial, which is why the question has you prove differentiability first. I believe your proof is valid for part (c), so just continue on with (d).