$\textbf {Problem:} \\\\ $
$\text{(a) Find integrating factor and then a general solution of ODE} \\$
$\begin{gather}
(1+xy+y^{2}) + (1+xy+x^{2})y' = 0
\end{gather}$
$\text{(b) Also, find a solution satisfying y(1) = 1}\\\\$
$\textbf{Solution: } \\\\$
$\text{(a):}\\\\$
$\text{We have:}$
$
\begin{gather}
\begin{aligned}
M_y = x+2y \\\\
N_x = y+2x
\end{aligned}
\end{gather}
$
$\text{Since} M_y \ne N_x \ \text{, thus equation is not exact.}$
$
\begin{gather}
\begin{aligned}
\frac{N_x - M_y}{M \cdot x - N \cdot y} &{}= \frac{y+2x-x-2y}{x+x^{2}y+xy^{2}-y-xy^{2}-x^{2}y} \\\\
&{} = \frac{x - y}{x - y} \\\\
&{} = 1
\end{aligned}
\end{gather}
$
$\text{Now, we have our integrating factor } \mu \ \text{computed as follow: }$
$
\begin{gather}
\begin{aligned}
\mu &{} = e^{\int {1} \, dxy} \\\\
&{} = e^{xy}
\end{aligned}
\end{gather}
$
$\text{Now multiplying the original equation with our integrating factor, we have: }$
$
\begin{gather}
\begin{aligned}
e^{xy} + e^{xy} \cdot xy + e^{xy} \cdot y^{2} + (e^{xy} + e^{xy} \cdot xy + e^{xy} \cdot x^{2})y' = 0
\end{aligned}
\end{gather}
$
$\text{Lastly, solve for general solution F(x,y): }$
$
\begin{gather}
\begin{aligned}
F(x,y) &{} = \int_{0}^{x} (e^{xy} + e^{xy} \cdot xy + e^{xy} \cdot y^{2}) \, dx + h'(y) \\\\
&{} = x \cdot e^{xy} + y \cdot e^{xy} = c
\end{aligned}
\end{gather}
$
$\text{(b):}\\\\$
$\text{By plugging in initial value condition, we have: }$
$
\begin{gather}
\begin{aligned}
y(1) = 1 \\\\
c = e + e = 2e\\\\
\implies x \cdot e^{xy} + y \cdot e^{xy} = 2e
\end{aligned}
\end{gather}
$
