$f$ is partially derivable if and only if $f$ is derivable in the direction of the standard unit vectors $e_{1},...,e_{n}$ $f$ is a function defined by, $f:U \to \mathbb{R}$ with $U \subset \mathbb{R^{n}}$ open. v $\in \mathbb{R^{n}}$ a unitvector then we say that $f$ is derivatible in a point $a \in U$ in the direction of v if the function $t\to f(a+t$v) is derivatble in $zero$
I want to proof that $f$ is partially derivable if and only if $f$ is derivable in the direction of the standard unit vectors $e_{1},...,e_{n}$ with $e_{i}=(0,...,0,1,0,....,0)$ with $1$ at place $i$.
My definition of partial derivatble is:
If $U\subset \mathbb{R^{n}}$ open and $f:U \to \mathbb{R}$ a function. We say that $f$ is partial derivatible in a point $(a_{1},...,a_{n}) \in U$ if for every $i:1,...,n$ the function $h_{i}: x\to f(a_{1},...,a_{i-1},x,a_{i+1},...,a_{n})$ is derivatible in $a_{i}$
I tried the following. Can someone check if it's correct and help me to understand my mistakes.
$=>$ $f$ is partial derivable. We need to prove that a function $t \to f(e_{i}+tv)$ is derivable in zero. $v$ is also a standard unit vector, with $v \in \mathbb{R^{n}}$. Let's call the previous function $g: \mathbb{R} \to \mathbb{R}$, then $g$'(t)=$f'(e_{i}+tv)v$. This exist beceause $f$ is partial derivative. Now $g'(0)=f'(e_{i})v$ which is also derivable beceause $f$ is partial derivative.
$<=$ if $f$ is derivative in all the directions of the standard unit vector, then we can write every other direction as a lineaire combination of the standard unit vectors. This implicate that $f$ will be partial derivative.
 A: This is false. For $n \ge 2$, the following three conditions on a function $f : \mathbb{R}^n \to \mathbb{R}$ and a point $p \in \mathbb{R}^n$ are in strictly increasing order by strength, and in particular are inequivalent:

*

*The partial derivatives $\frac{\partial f}{\partial x_i}$ exist at $p$.

*The directional derivatives $\frac{\partial f}{\partial v}$ exist for every $v \in \mathbb{R}^n$ at $p$.

*$f$ is differentiable at $p$.

(I'm not sure whether by "partially derivable" you mean 2 or 3, but the statement is false either way.)
Your argument for $\Leftarrow$ does not work; you've just assumed that the partial derivative exists which is exactly what you were trying to prove. It is true that if $f$ is differentiable then every directional derivative is a linear combination of partial derivatives $\frac{\partial f}{\partial x_i}$, but you can't assume that here.
ancientmathematician's nice counterexample in the comments of the function $f$ which is equal to $1$ on the coordinate axes and $0$ otherwise shows that 1 does not imply 2 (which also shows that 1 does not imply 3); the partial derivatives $\frac{\partial f}{\partial x_i}$ exist at the origin and are equal to $0$, but the directional derivatives in all other directions fail to exist at the origin.
You can find a counterexample showing that 2 does not imply 3 at this MO question, namely the function
$$f(x, y) = 
  \begin{cases}
        0 & \text{for } (x,y)=(0,0) \\
   \frac{x^3}{x^2+y^2}       & \text{for } (x,y) \neq (0,0)
  \end{cases}$$
which is also a counterexample showing that "the directional derivatives exist for every $v$ at every point" does not imply "$f$ is differentiable at every point."
