(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 $(0,0)$.
Going with the hints in comments...
$$\lim_{(x,y) \to (0,0)} \frac{\big|2 \sin x - 2x + x^2 y \sin \big(\frac{1}{x^2 + y^4}\big)\big|}{\sqrt{x^2 + y^4}} \leq \lim_{(x,y) \to (0,0)} \Big|\frac{2 \sin x - 2x}{x}\Big| + \lim_{(x,y) \to (0,0)} \big|xy \sin \big(\frac{1}{x^2 + y^4}\big)\big|$$.
By noting that $\frac{\sin x - x}{x} < 0$ for any $x \neq 0$, we can get rid of the absolute value in the first limit. It is then easily seen to be zero. We use the fact that $0 \leq |\sin( \cdot )| \leq 1$ to obtain the following upper bound to the second limit: $\lim_{(x,y) \to (0,0)} |xy|$. While it may be obvious that this bound is $0$, it is in my opinion more rigourous to use the $\varepsilon,\delta$-definition for a quick proof.
Let $\varepsilon > 0$ and take $\delta = \sqrt{\varepsilon} > 0$. Choose $(x,y) \in \mathbb{R}^2$ such that $||(x,y)|| = \sqrt{x^2 + y^2} < \delta$. Then we surely have $|x| < \delta$ and $|y| < \delta$ and, therefore, $|xy| < \delta^2 = \varepsilon$. This proves that $\lim_{(x,y) \to (0,0)} |xy| = 0$.
We have now bounded the desired limit above by 0. But since the expression inside the limit itself is always greater than or equal to 0, the limit must be 0. This proves that $f$ is totally differentiable in $(0,0)$.
Best Answer
Here is a proof that $$\lim_{(x,y) \rightarrow (0,0)} \frac{f(x,y) - f(0,0)}{\sqrt{x^2+y^2}}=0.$$
Let $\varepsilon>0$. Take $\delta=\varepsilon$, let $(x,y) \in \mathbb{R}^2\backslash \{(0,0)\}$ such that $||(x,y) - (0,0)|| \leq \delta$.
\begin{align} \left|\frac{f(x,y) - f(0,0)}{\sqrt{x^2+y^2}} - 0\right| &= \left|\frac{x^3y}{(x^2 + y^4) \sqrt{x^2+y^2}}\right| \\ &= \left| \frac{x^2}{x^2 + y^4} \right| \left| \frac{x}{\sqrt{x^2+y^2}} \right| \left| y \right| \\ &\leq \left| y \right| \leq \delta = \varepsilon. \end{align}
So, $f$ is differentiable at $(0,0)$.