(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 $(0,0)$.
As indicated by @levap in the comments, you are likely confusing the (right-hand side, RHS) derivative at $0$ with the (RHS) limit of the derivative. I will just add details to clarify. Assuming you mean $f(0)=0$.
The derivative at $0$ is by definition
$\displaystyle\lim\limits_{h\to0}\frac{f(0+h)-f(0)}h = \lim\limits_{h\to0}\frac{h^2\sin(\frac1h)-0}h =
\lim\limits_{h\to0}h\sin(\frac1h)=0$.
Similarly the RHS derivative at $0$ is
$\displaystyle\lim\limits_{h\to0^+}\frac{f(0+h)-f(0)}h = \lim\limits_{h\to0^+}\frac{h^2\sin(\frac1h)-0}h =
\lim\limits_{h\to0^+}h\sin(\frac1h)=0$,
and the LHS derivative at $0$ is
$\displaystyle\lim\limits_{h\to0^-}\frac{f(0+h)-f(0)}h = \lim\limits_{h\to0^-}\frac{h^2\sin(\frac1h)-0}h =
\lim\limits_{h\to0^-}h\sin(\frac1h)=0$.
On the other hand when $x\not=0$ we could use differentiation formulas (apart from the limit definition of the derivative, which we could also use but it would be unnecessary), to obtain that
$f'(x)=2x\sin(\frac1x)-\cos(\frac1x)$, $x\not=0$.
Now, it is true that the limits $\lim\limits_{x\to0}f'(x)$, as well as $\lim\limits_{x\to0^+}f'(x)$ and $\lim\limits_{x\to0^-}f'(x)$ do not exist, that is the derivative $f'(x)$ has no limit (no RHS limit, no LHS limit) as $x\to0$, but this is a different matter than the derivative (the RHS derivative, the LHS derivative) of $f$ at $0$. The latter, as we saw earlier, exists and equals to $0$.
To illustrate, I enclose a couple of graphs (different zoom). $f(x)$ is shown in black, squeezed in between grey $x^2$ an $-x^2$. It should be clear from the picture that the tangent line at $0$ exists and has slope $0$. On the other hand in every neighborhoos of $0$ the slope of $f$ varies between $\approx\pm1$. That is, the dirivative $f'(x)$, shown in red, varies between $\approx\pm1$ (when $x$ is close to $0$ but different from $0$), and hence the limit of the derivative does not exist, as $x\to0$.
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Best Answer
The statement the book enunciates is completely wrong. The easiest counterexample for it would be any constant function $f:\mathbb{R}\to \mathbb{R}$. As you suggest, the function: $$f(x) = |x^3| = \begin{cases} x^3 & , x\geq 0 \\ - x^3 & , x<0 \end{cases} $$ , is both continuous and differentiable in the whole real axis.