There are no directional derivatives in nearly all directions. Consider, in particular, along the line $y=x$. $f(x,y)$ is a constant times the absolute value function.
When a function of two variables is differentiable, then there is a tangent plane to the surface $z=f(x,y)$, and there are directional derivatives in all directions. This one doesn't have directional derivatives except in two directions, and there's no tangent plane to the surface $z=f(x,y)$.
(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 $(0,0)$.
Best Answer
We have $${ f }_{ x }^{ \prime }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } }{ x } } =1\\ \\ { f }_{ y }^{ \prime }\left( 0,0 \right) =\lim _{ y\rightarrow 0 }{ \frac { f\left( 0,y \right) -f\left( 0,0 \right) }{ y } } =\lim _{ y\rightarrow 0 }{ \frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } }{ y } } =0\\ f\left( x,y \right) -f\left( 0,0 \right) =\frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } =x+\left( \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } -x \right) ={ f }_{ x }^{ \prime }\left( 0,0 \right) x+{ f }_{ y }^{ \prime }\left( 0,0 \right) y+\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } $$ where $\alpha \left( x,y \right) =\frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } -x }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } =\frac { -2x{ y }^{ 2 } }{ \left( x^{ 2 }+y^{ 2 } \right) \sqrt { x^{ 2 }+y^{ 2 } } } $ however when $n\rightarrow \infty $ $$\alpha \left( \frac { 1 }{ n } ,\frac { 1 }{ n } \right) =\frac { -\frac { 1 }{ { n }^{ 3 } } }{ \frac { 1 }{ { n }^{ 3 } } \sqrt { 2 } } =-\frac { 1 }{ \sqrt { 2 } } $$ which shows $$\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \neq o\left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right) $$