Differentiability of function of two variables at $(1,0)$

Define $$F:\mathbb{R}^2\to\mathbb{R}$$ by $$F(x,y)=y^2+\sqrt{y^2x^4}+x$$.

Determine if F is differentiable at (1,0) or not.

To solve this problem, I have tried to applied $$\lim_{x \to a} \frac{||F(x)-F(a)-T(x-a)||}{||x-a||}$$ to figure out whether the limit is equal to zero or not.

Then I got $$\lim_{(x,y) \to (1,0)}\frac{y^2+yx^2-y}{\sqrt{(x-1)^2+y^2}}$$ , the problem is I do not know how to solve such a limit or I used a wrong method to determine the differentiability.

In the other sub-part of the question, I have already solved the function is differentiable at (1,1) and (0,0).

• $F$ is not a vector valued function. What is $T$? – uniquesolution Mar 6 at 18:19
• I thought the existence of partial derivatives at a point were a necessary condition for differentiability at that point. At $(0,0)$ they don't exist do they? so how did you prove it is differentiable at $(0,0)$? They don't seem to exist at $(1,0)$ either. Don't take my word, I'm also a student so hopefully someone will answer this soon (and as mentioned above, F is not a vector valued function since it takes values in reals - it is real valued.) – Displayname Mar 6 at 18:34
• @Displayname The partials do exist at $(0,0).$ – zhw. Mar 6 at 18:37
• @zhw im a bit confused now...so what would the partial of $F$ w.r.t y be at (0,0)? – Displayname Mar 6 at 18:42
• @Displayname $F(0,y)=y^2,$ so the partial of $F$ wrt $y$ is $0$ at the origin. – zhw. Mar 6 at 19:01

Hint: $$F(x,y)=y^2+ |y|x^2 +x.$$ Does $$\partial F/\partial y (1,0)$$ exist?