For $f: R^2 \to R$ to be diffrentiable in point $p = (p_x;p_y)$ it has to satisfy for aribitrary $\theta$
$$1)\space\space\lim_{r \to 0} \frac{f(p_x + rcos(\theta);p_y + rsin(\theta)) - f(p_x;p_y) - \cfrac{df}{dx}(p_x;p_y)rcos(\theta) - \cfrac{df}{dy}(p_x;p_y)rsin(\theta)}{r} = 0$$
We can estimate $f(p_x + rcos(\theta);p_y + rsin(\theta))$ using value $f(p_x;p_y)$ and partial derivatives of $f$ in a following way.
$$2)\space\space f(p_x + rcos(\theta);p_y + rsin(\theta)) = f(p_x;p_y) + \frac{df}{dx}(p_x;p_y)rcos(\theta) + \frac{df}{dy}(p_x + rcos(\theta);p_y)rsin(\theta) + E_x + E_y$$ Where $E_x$ and $E_y$ are errors of approximation and based on definition of derivative $$\lim_{r \to 0}{\frac{E_x}{r}} =0 $$ and $$\lim_{r \to 0}{\frac{E_y}{r}} =0 $$ Substituting 2 to 1 and I end up with $$\lim_{r \to 0} \frac{rsin(\theta)(\cfrac{df}{dy}(p_x+rcos(\theta);p_y)-\cfrac{df}{dy}(p_x;p_y))}{r} + \lim_{r \to 0}{\frac{E_x}{r}} + \lim_{r \to 0}{\frac{E_y}{r}}$$ Now if $\frac{df}{dy}$ is continuous at p the first limes is 0 and limit of whole expression is also 0 implying $f$ is frechet diffrentiable at p. However I only used continuity of one partial derivative (other just has to exist), while known theorem states that all partial derivatives should be continous to imply diffrentiability. Do I make a mistake somewhere along the way? It would be great if someone could provide couterexample of non diffrentiable function with one continous partial derivative. Thank you in advance.
No, you're not making a mistake
It is a general result that if $f: \mathbb R^n \to \mathbb R$ has only $n-1$ continuous partial derivatives at $x \in \mathbb R^n$, then $f$ is Fréchet differentiable at $x$. Unfortunately this stronger result is rarely mentioned in the litterature.