First, note that $g'(x)=1$ for $x>1$.
On the interior of each of the intervals $(-\infty,-1)$, $[-1,1]$, and $(1,\infty)$ $g$ is differentiable since each component function is. The only question is what happens at the endpoints of these intervals.
At $x=-1$, the value of both component functions is $1$ and the derivative of both component functions is $2$. This means we get
$$
\lim_{h\to0^-}\frac{g(-1+h)-g(-1)}{h}=-2\tag{1}
$$
Because $x^2=-1-2x$ at $x=-1$, we can use $g(x)=-1-2x$ for the computation of $(1)$.
Furthermore, we get
$$
\lim_{h\to0^+}\frac{g(-1+h)-g(-1)}{h}=-2\tag{2}
$$
using $g(x)=x^2$ for the computation of $(2)$.
Since the derivatives computed in $(1)$ and $(2)$ are the same, we get that $g(x)$ is differentiable at $x=-1$.
At $x=1$, the value of both component functions is $1$, however, the derivative of $x^2$ is $2$ and the derivative of $x$ is $1$. This means that
$$
\lim_{h\to0^-}\frac{g(1+h)-g(1)}{h}=2\tag{3}
$$
using $g(x)=x^2$ for the computation of $(3)$.
However, we get
$$
\lim_{h\to0^+}\frac{g(1+h)-g(1)}{h}=1\tag{4}
$$
Because $x=x^2$ at $x=1$, we can use $g(x)=x$ for the computation of $(4)$.
Since the derivatives computed in $(3)$ and $(4)$ are different, $\lim\limits_{h\to0}\frac{g(1+h)-g(1)}{h}$ does not exist, and therefore $g(x)$ is not differentiable at $x=1$.
(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
Because $f$ is continuous at $x=1$, $l + k + m = k + l \Rightarrow m = 0$
Because $f$ is also differentiable at $x=1$, \begin{align} L = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} \end{align} exists. \begin{align} \tag{1} L_-\lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^-} \frac{k(1+h)^2+l-k-l}{h} = \lim_{h \to 0^-} \frac{2kh+kh^2}{h} = 2k \newline \tag{2} L_+=\lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^-} \frac{l(1+h)^2+k+kh-k-l}{h} = \lim_{h \to 0^-} \frac{2lh+lh^2+kh}{h} = 2l + k \end{align} For the limit $L$ to exist, $L_- = L_+ \Rightarrow 2k = 2l+k \Rightarrow k = 2l$
Then, $k-2l+m = 2l - 2l + 0 = 0$