(a) it is a compostion of diferentiable functions, then it is differentiable, and contious and the partial derivative exist in $(0,0)$.
(b) It is continuous,and we have that the partial derivative are
$f_x(0,0)=\displaystyle\lim_{t\to 0}\frac{f(t,0)-f(0,0)}{t}=
\lim_{t\to 0}\frac{\sqrt{|t0|}-\sqrt{|0\times0|}}{t}=0$, and
$f_y(0,0)=\displaystyle\lim_{t\to 0}\frac{f(0,t)-f(0,0)}{t}=
\lim_{t\to 0}\frac{\sqrt{|t0|}-\sqrt{|0\times0|}}{t}=0$. however it is not differentiable since $\lim_{t\to0^+}\frac{f(t,t)-f(0,0)}{t}=1$ and $\lim_{t\to0^-}\frac{f(t,t)-f(0,0)}{t}=-1$,then it is not diferentiable.
(c) It is continuous because it is composition of continuous functions,but
$f_x(0,0)=\displaystyle\lim_{t\to 0^+}\frac{f(t,0)-f(0,0)}{t}=
\lim_{t\to 0^+}\frac{1 − \sin\sqrt{t^2 + 0^2}- (1 − \sin\sqrt{0})}{t}=-1$ but
$f_x(0,0)=\displaystyle\lim_{t\to 0^-}\frac{f(t,0)-f(0,0)}{t}=
\lim_{t\to 0^-}\frac{1 − \sin\sqrt{t^2 + 0^2}- (1 − \sin\sqrt{0})}{t}=1$
and the partial $f_x$ is not defined in $(0,0)$, analogously for $f_y(0,0)$, both does not exist.
Then it is not differentiable, because a differentiable function the elimites above should exist.
(d) it is not continuous, because $t>0$ then $(t,t)\to 0$ then $f(t,t)=1/2\neq 0=f(0,0)$. And it is not differentiable since it is not continuous. However
$f_x(0,0)=\displaystyle\lim_{t\to 0}\frac{f(t,0)-f(0,0)}{t}=
\lim_{t\to 0^+}\frac{\dfrac{t0}{t^2+0^2}-0}{t}=0$ and
$f_y(0,0)=\displaystyle\lim_{t\to 0}\frac{f(0,t)-f(0,0)}{t}=
\lim_{t\to 0^+}\frac{\dfrac{t0}{t^2+0^2}-0}{t}=0$.
(e) It is clearly not continuous, hence not differentiable at $(0,0)$, but
$f_x=\displaystyle\lim_{t\to0}\frac{f(x+t,y)-f(x,t)}{t}=0$ and
$f_y=\displaystyle\lim_{t\to0}\frac{f(x,y+t)-f(x,t)}{t}=0$, are defined in $(0,0)$
(f)It is not continuous since $\lim_{t\to 0}f(2t,t)=\lim_{t\to0}\dfrac{4t^2-t^2}{4t^2+t^2}=\frac{3}{5}\neq f(0,0)$, hence it is not differentiable in $(0,0)$.
$f_x(0,0)=\displaystyle\lim_{t\to0^+}\frac{f(x+t,y)-f(x,t)}{t}
=\lim_{t\to 0^+}\frac{\dfrac{t^2-0^2}{t^2+0^2}-0}{t}=+\infty
$ analogously for $f_y(0,0)$, both are note defined in $(0,0)$.
Here's one option. Write $(x,y)$ in polar form: $x = r\cos(\theta)$, $y = r\sin(\theta)$. You get:
$$u(r,\theta) = \frac{r^3\cos^{3}(\theta) - 3r^3\cos(\theta)\sin^{2}(\theta)}{r^2}$$
$$u(r,\theta) = r[\cos^{3}(\theta) - 3\cos(\theta)\sin^{2}(\theta)]$$
Since $\cos^{3}(\theta) - 3\cos(\theta)\sin^{2}(\theta)$ is continuous, it is bounded. Meaning there exists some $M > 0$ such that $-M < \cos^{3}(\theta) - 3\cos(\theta)\sin^{2}(\theta) < M$.
Then you have $-Mr < u(r,\theta) < Mr$. By the squeeze theorem:
$$\lim_{(x,y) \rightarrow (0,0)}u(x,y) = \lim_{r \rightarrow 0^{+}}u(r,\theta) = 0$$
Best Answer
You can notice that for $y \gt 0$
$$\begin{aligned}\vert f(x,y) \vert &= \left\vert \frac{x^2y}{x^2+\sqrt{y}} \right\vert\\ &\le \left\vert \frac{x^2y}{x^2} \right\vert = \vert y \vert \end{aligned}$$ and that the exact same inequality is also satisfied for $y \le 0$ as in that case $f$ vanishes.
As for any $u \in \mathbb R$ $\lim\limits_{(x,y) \to (u,0)} \vert y \vert = 0$, you can conclude that $f$ is continuous at $(u,0)$ as $f(u,0) = 0$.
Also, $f$ is continuous at $(u,v)$ with $v \neq 0$ using continuity of composition of continuous maps.
Finally, $f$ is continuous everywhere.