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I have to check whether the function is able to be differentiated on M(0, 0), and find partial derivatives $f_x'(0, 0), f'_y(0, 0)$. Is it correct?

Let $z = {x}+{y}+\sqrt{\mid{xy}\mid}$. By definition of partial derivative, $$\frac{\partial{z}}{\partial{x_k}} = \lim_{\Delta{x}\to0}{\frac{f(x_1,\dots,x_k+\Delta{x}_1,\dots,x_n)-f(x_1,\dots,x_k,\dots,x_n)}{\Delta{x}}}$$

Therefore, we calculate the partial derivative with respect to $x$: $$\Delta{z} = f(x_0+\Delta{x}, y_0) - f(x_0, y_0)$$ $$\Delta{z} = (x_0+\Delta{x}+y_0+\sqrt{\mid(x_0+\Delta{x})y\mid}) - (x_0+y_0+\sqrt{\mid x_0y_0\mid}) $$ $$(x_0,y_0)=(0, 0)\rightarrow\Delta{x}+\sqrt{0(\mid{0+\Delta{x}}\mid)}-0=\Delta{x}$$ $$\lim_{\Delta{x}\to0}\frac{\Delta{x}}{\Delta{x}} = 1.$$

Then, with respect to $y$: $$\Delta{z} = f(x_0, y_0 + \Delta{y}) - f(x_0, y_0)$$ $$\Delta{z} = (x_0+\Delta{y}+y_0+\sqrt{\mid(y_0+\Delta{y})x\mid}) - (x_0+y_0+\sqrt{\mid x_0y_0\mid}) $$ $$(x_0,y_0)=(0, 0)\rightarrow\Delta{y}+\sqrt{0(\mid{0+\Delta{y}}\mid)}-0=\Delta{y}$$ $$\lim_{\Delta{y}\to0}\frac{\Delta{y}}{\Delta{y}} = 1.$$

Thus, the partial derivatives $f'_x(0, 0)$, $f'_y(0, 0)$ do exist and equal $1$.

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Yes it is correct, indeed by another notation we have that

$$f_x(0,0)=\lim_{h\to 0} \frac{{(0+h)}+{0}+\sqrt{\mid{(x+h)0}\mid}-0}{h}=\lim_{h\to 0} \frac{h}{h}=1$$

and $f_y(0,0)=f_x(0,0)$ by symmetry.

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That part is OK.

Now, to see that $f$ is NOT differentiable at $(0,0)$, remember that if $f$ is differentiable at $(0,0)$, then for any direction $\vec v$ such that $\|\vec v\|=1$ it is true that $$f'_{\vec v}(0,0)=\langle \nabla f(0,0),\vec v\rangle.$$ But you proved that $\nabla f(0,0)=(1,1)$, and you can use the definition to calculate $f'_{\vec v}(0,0)$ for any other direction, such as $(\frac1{\sqrt{2}},\frac1{\sqrt{2}})$, $(\frac35,-\frac45)$, etc., and then do the inner product on the right hand side to see that the formula does not hold. This can only be the case if $f$ is not differentiable at $(0,0)$.

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