Answer
(a)
$$
f(0)=0
$$
(b)
$$
f^{\prime}(0)=1
$$
(c)
$$
f^{\prime } (x)=1+x^{2}
$$
Work Step by Step
Since
$$
f(x+y)=f(x)+f(y)+x^{2}y+xy^{2} \quad\quad (*)
$$
(a) To find $f(0)$, put $x=0$ and $y=0$ into the equation above:
$$
f(0+0)=f(0)=f(0)+f(0)+0^{2} .0+0 .0^{2} =2 f(0)
$$
$\Rightarrow $
$$
f(0)=0
$$
(b) To find $f^{\prime } (0)$:
$$
\begin{aligned} f^{\prime}(0) &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\
&=\lim _{h \rightarrow 0} \frac{\left[f(0)+f(h)+0^{2} h+0 h^{2}\right]-f(0)}{h} \\
&=\lim _{h \rightarrow 0} \frac{f(h)}{h} \\
&=\lim _{x \rightarrow 0} \frac{f(x)}{x} \\
&=1
\end{aligned}
$$
(c) To find $f^{\prime } (x)$
$$
\begin{aligned}
f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{\left[f(x)+f(h)+x^{2} h+x h^{2}\right]-f(x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{f(h)+x^{2} h+x h^{2}}{h} \\
&=\lim _{h \rightarrow 0}\left[\frac{f(h)}{h}+x^{2}+x h\right] \\
&=1+x^{2}
\end{aligned}
$$
