Let $f(x_1, x_2, \ldots x_n)$ be a real-valued function in $\mathbb{R}^n$, $n \in \mathbb{N}$. Assume that for any integer $i \in [1,n]$, and for any collection of real values of $(x_j)_{j \neq i}$, the function $$ F(x_i) = f(x_1, \ldots, x_i, \ldots x_n) $$ is differentiable in $\mathbb{R}$ (which means, we keep the coordinates $x_j$ for $j \neq i$ fixed and we consider $f$ as a function only of $x_i$).
Does it necessarily mean that $f(x_1, \ldots, x_n)$ is differentiable in $\mathbb{R}^n$?
Possibly, the simplest counterexample is
$$f(x,y) = \begin{cases}\frac{xy}{x^2 + y^2}, \quad x^2 + y^2 > 0 \\ 0, \quad\,\,\,\,\,\quad x = y = 0 \end{cases}$$
where $\displaystyle \left.\frac{\partial f}{\partial x }\right|_{(0,0)} = \left.\frac{\partial f}{\partial y }\right|_{(0,0)} = 0,$ but $f$ is not differentiable at $(0,0)$ since it fails to be continuous.
Note that
$$\lim_{(x,y) \to (0,0), x = 0}f(x,y) = 0 \neq \frac{1}{2} = \lim_{(x,y) \to (0,0), x = y}f(x,y), $$
but
$$\left.\frac{\partial f}{\partial x }\right|_{(0,0)} = \lim_{h \to 0} \frac{f(0+h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{0 - 0}{h} = 0$$