Prove existence of partial derivatives Let $f:\mathbb{R}^n\to\mathbb{R}$ be such that $f(\lambda x)=\lambda f(x)$ for any $x \in \mathbb{R}^n$ and any $\lambda\in\mathbb{R}$.
I'm looking to prove that the partial derivatives of $f$ exist, however I don't have much of an idea on how to begin. I took a guess based on what I can work with and thought that if the partial derivatives $\frac{\partial f}{\partial x_i}$ exist, they would have to satisfy the following equation (1):
$$\frac{\partial f}{\partial x_i}(\lambda x_1,\cdots,\lambda x_i,\cdots,\lambda x_n)=\lambda\frac{\partial f}{\partial x_i}(x_1,\cdots,x_i,\cdots,x_n)$$
so I went and calculated the partials:
$$\lim_{h\to 0}\frac{f(\lambda x_1,\cdots,\lambda (x_i+h),\cdots,\lambda x_n)-f(\lambda x_1,\cdots,\lambda x_i,\cdots,\lambda x_n)}{h}$$
using the hypothesis, we certainly have that the limit above is equal to
$$\lim_{h\to 0}\lambda \cdot\frac{f(x_1,\cdots,(x_i+h),\cdots,x_n)-f(x_1,\cdots,x_i,\cdots,x_n)}{h}$$
and so equation (1) is satisfied. I feel like this assumes existence of partial derivatives, so it's obviously not what I'm trying to prove. But I haven't thought about any other possible approach...
 A: We show in general such a function does not have well defined partials at every point of $\mathbb{R}^{n}$.
Take $n = 2$, $f:\mathbb{R^2} \to \mathbb{R}$ given by 
$$f(x,y) = \begin{cases} \frac{x^2 + y^2}{x} &\text{ if } x \neq 0 \\
                   0 &\text{ if } x = 0. \end{cases}$$
Let $\lambda \in \mathbb{R}$. If $\lambda = 0$ then $f(\lambda x, \lambda y) = f(0,0) = 0 = \lambda f(x,y)$. Consider $\lambda \neq 0$. If $x \neq 0$ then $\lambda x \neq 0$ so 
$$f(\lambda x, \lambda y) = \frac{(\lambda x)^2 + (\lambda y)^2}{\lambda x} = \frac{\lambda^2}{\lambda}f(x,y) = \lambda f(x,y)$$.
If $x = 0$, then $\lambda x = 0$ so 
$$f(\lambda x, \lambda y) = 0 = \lambda 0 = \lambda f(x,y)$$. 
Thus for any $\lambda \in \mathbb{R}$ and $(x,y) \in \mathbb{R}^2$, $f(\lambda x, \lambda y) = \lambda f(x,y)$. 
Next I claim that $f$ does not have a partial at $(0,1)$ with respect to $x$. We have 
$$\lim_{h \to 0}\frac{f(0 + h, 1) - f(0,1)}{h} = \lim_{h \to 0} \frac{h^2 + 1}{h}  = \lim_{h \to 0} h + \frac{1}{h}\to \infty.$$
Thus in general a function with this property will not have partial derivatives at every point. 
