The equation cam be written as

\(\displaystyle{\left({2}{x}{y}-{9}{x}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({2}{y}+{x}^{{2}}+{1}\right)}{\left.{d}{y}\right.}={0}\)

The equation is of the form

\(\displaystyle{M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}={0}\)

\(\displaystyle{M}{y}={2}{x}\)

\(\displaystyle{N}{x}={2}{x}\)

The equation is exact since My=Nx

Assume that F(x,y) is the solution for the equation

\(\displaystyle{F}{x}={M}\)

\(\displaystyle={2}{x}{y}-{9}{x}^{{2}}\)

\(\displaystyle{F}=\int{2}{x}{y}-{9}{x}^{{2}}{\left.{d}{x}\right.}\)

\(\displaystyle{F}={x}^{{2}}{y}-{3}{x}^{{2}}+{g{{\left({y}\right)}}}\)

Take the derivative for F with respect to y

\(\displaystyle{F}{y}={x}^{{2}}+{g}'{\left({y}\right)}\)

\(\displaystyle{2}{y}+{x}^{{2}}+{1}={g}'{\left({y}\right)}\)

\(\displaystyle{g}'{\left({y}\right)}={2}{y}+{1}\)

\(\displaystyle{g}'{\left({y}\right)}=\int{2}{y}+{1}{\left.{d}{y}\right.}\)

\(\displaystyle{g}'{\left({y}\right)}={y}^{{2}}+{y}\)

Hence

\(\displaystyle{x}^{{2}}{y}-{3}{x}^{{3}}+{y}^{{2}}+{y}={C}\)

\(\displaystyle{y}^{{2}}+{\left({x}^{{2}}-{1}\right)}{y}-{3}{x}^{{2}}={C}\)

It is known that

\(\displaystyle{y}{\left({0}\right)}=-{3}\)

\(\displaystyle{9}-{3}={c}\)

\(\displaystyle{c}={6}\)

So the solution for the problem is

\(\displaystyle{y}^{{2}}{\left({x}^{{2}}-{1}\right)}{y}-{3}{x}^{{2}}-{6}={0}\)

\(\displaystyle{\left({2}{x}{y}-{9}{x}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({2}{y}+{x}^{{2}}+{1}\right)}{\left.{d}{y}\right.}={0}\)

The equation is of the form

\(\displaystyle{M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}={0}\)

\(\displaystyle{M}{y}={2}{x}\)

\(\displaystyle{N}{x}={2}{x}\)

The equation is exact since My=Nx

Assume that F(x,y) is the solution for the equation

\(\displaystyle{F}{x}={M}\)

\(\displaystyle={2}{x}{y}-{9}{x}^{{2}}\)

\(\displaystyle{F}=\int{2}{x}{y}-{9}{x}^{{2}}{\left.{d}{x}\right.}\)

\(\displaystyle{F}={x}^{{2}}{y}-{3}{x}^{{2}}+{g{{\left({y}\right)}}}\)

Take the derivative for F with respect to y

\(\displaystyle{F}{y}={x}^{{2}}+{g}'{\left({y}\right)}\)

\(\displaystyle{2}{y}+{x}^{{2}}+{1}={g}'{\left({y}\right)}\)

\(\displaystyle{g}'{\left({y}\right)}={2}{y}+{1}\)

\(\displaystyle{g}'{\left({y}\right)}=\int{2}{y}+{1}{\left.{d}{y}\right.}\)

\(\displaystyle{g}'{\left({y}\right)}={y}^{{2}}+{y}\)

Hence

\(\displaystyle{x}^{{2}}{y}-{3}{x}^{{3}}+{y}^{{2}}+{y}={C}\)

\(\displaystyle{y}^{{2}}+{\left({x}^{{2}}-{1}\right)}{y}-{3}{x}^{{2}}={C}\)

It is known that

\(\displaystyle{y}{\left({0}\right)}=-{3}\)

\(\displaystyle{9}-{3}={c}\)

\(\displaystyle{c}={6}\)

So the solution for the problem is

\(\displaystyle{y}^{{2}}{\left({x}^{{2}}-{1}\right)}{y}-{3}{x}^{{2}}-{6}={0}\)