Let $f: \mathbb{R}^2\to \mathbb{R}$ be defined by setting $f(x,y)=|xy|^{1/2}$
(a) For which vectors $u\ne 0$ does $f'(0; u)$ exist? Evaluate it when it exists.
(b) Do $D_1f$ and $D_2f$ exist at $0$?
(c) Is $f$ differentiable at $0$?
(d) Is $f$ continuous at $0$?
I have thought a lot about this problem:
For (a), be $u\neq 0, u:=(h,k)$, then $\lim_{t\to 0}\frac{f(0+tu)-f(0)}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{|t^2hk|^{1/2}}{t}=|hk|^{1/2}\lim_{t\to 0}\frac{|t|}{t}$ and as $\lim_{t\to 0}\frac{|t|}{t}$ does not exist then necessarily $hk=0$.
For (b), we have that $\lim_{t\to 0}\frac{f(0+te_1)-f(0)}{t}=\lim_{t\to 0}\frac{f(t,0)}{t}=\lim_{t\to 0}0=0$ and so $D_1f(0,0)=0$. Plus $\lim_{t\to 0}\frac{f(0+te_2)-f(0)}{t}=\lim_{t\to 0}\frac{f(0,t)}{t}=\lim_{t\to 0}0=0$, then $D_2f(0,0)=0$.
For (c), $f$ is not differentiable in $(0,0)$ since the directional derivative of $f$ in $a$ with respect to the vector $u=(1,1)$ does not exist since $\lim_{t\to 0}\frac{f(t,t)-f(0)}{t}=\lim_{t\to 0}\frac{|t^2|^{1/2}}{t}=\lim_{t\to 0}\frac{|t|}{t}$ and we know that $\lim_{t\to 0}\frac{|t|}{t}$ does not exist.
For (d), $f$ is continuous in $(0,0)$ because $\lim_{(x,y)\to (0,0)}f(x,y)=\lim_{(x,y)\to (0,0)}|xy|^{1/2}=0$ and $f(0,0)=0$.
Is this proof of continuity okay? Could anyone help me, please? Thank you.