The vector field \(\displaystyle{F}=\Pi+{Q}{j}\) is convervative if and only if

\(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{y}\right.}}}}={\frac{{{d}{Q}}}{{{\left.{d}{x}\right.}}}}\)

The given vector field is \(\displaystyle{F}{\left({x},{y}\right)}={\left({3}{x}^{{2}}-{2}{y}^{{2}}\right)}{i}+{\left({4}{x}{y}+{3}\right)}{j}\)

Here \(\displaystyle{P}={3}{x}^{{2}}-{2}{y}^{{2}}\), Therefore \(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{y}\right.}}}}=-{4}{y}\)

And \(\displaystyle{Q}={4}{x}{y}+{3}\), Therefore \(\displaystyle{\frac{{{d}{Q}}}{{{\left.{d}{x}\right.}}}}={4}{y}\)

\(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{y}\right.}}}}\ne{\frac{{{d}{Q}}}{{{\left.{d}{x}\right.}}}}\)

Therefore F is NOT conservative.

\(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{y}\right.}}}}={\frac{{{d}{Q}}}{{{\left.{d}{x}\right.}}}}\)

The given vector field is \(\displaystyle{F}{\left({x},{y}\right)}={\left({3}{x}^{{2}}-{2}{y}^{{2}}\right)}{i}+{\left({4}{x}{y}+{3}\right)}{j}\)

Here \(\displaystyle{P}={3}{x}^{{2}}-{2}{y}^{{2}}\), Therefore \(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{y}\right.}}}}=-{4}{y}\)

And \(\displaystyle{Q}={4}{x}{y}+{3}\), Therefore \(\displaystyle{\frac{{{d}{Q}}}{{{\left.{d}{x}\right.}}}}={4}{y}\)

\(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{y}\right.}}}}\ne{\frac{{{d}{Q}}}{{{\left.{d}{x}\right.}}}}\)

Therefore F is NOT conservative.