(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 $(0,0)$.
The continuity of the partial derivatives $f_x$, $f_y$ at $(x_0,y_0)$is a sufficient, but not a necessary condition for differentiability of $f$ at $(x_0,y_0)$.
In your example both $f_x$ and $f_y$ are continuous on the open set $\dot{\mathbb R}^2:={\mathbb R}^2\setminus\{(0,0)\}$; whence $f$ is differentiable at all points of $\dot{\mathbb R}^2$.
It remains to test the point ${\bf 0}:=(0,0)$. According to the definition of $f$ we have
$$f({\bf 0}+{\bf h})-f({\bf 0})=|{\bf h}|^2\sin{1\over |{\bf h}|^2}\qquad({\bf h}\ne{\bf 0})$$
and therefore
$${f({\bf 0}+{\bf h})-f({\bf 0})-{\bf 0}\cdot{\bf h}\over|{\bf h}|}=|{\bf h}|\sin{1\over |{\bf h}|^2}\to{\bf 0}\qquad({\bf h}\to{\bf 0})\ .\tag{1}$$
Here $\cdot$ denotes the scalar product in ${\mathbb R}^2$. The last formula proves $df({\bf 0})=0$, and a fortiori $\nabla f({\bf 0})={\bf 0}$. It follows that $f$ is differentiable on all of ${\mathbb R}^2$.
Note that computing $f_x(0,0)=f_y(0,0)=0$ does not suffice. We need a formula of type $(1)$ to assure differentiability at ${\bf 0}$.
Best Answer
For (a) you can follow Hans Lundmark’s suggestion. Alternatively, you can use the fact that $$0\le (x-y)^2 = x^2-2xy+y^2$$ to deduce that $2xy\le x^2+y^2$, from which it’s easy to bound the function.
For (c), what happens if you approach the origin along the line $y=x$? Then for $x\ne 0$ your function is just $$f(x) =\frac{x^2}{\sqrt{2x^2}},$$ whose derivative is pretty easy to investigate. What if you approach the origin along the $x$-axis?