Use implicit differentiation

Need to know that: \(\displaystyle{\left(\frac{\partial}{\partial}{x}\right)}{y}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={y}′\)

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}{\left({3}{y}+{2}{x}\times{\ln{{\left({y}\right)}}}={y}^{{{4}}}+{x}\right.}\) (expand by chain rule/product rule)

\(3y'+2\times\ln(y)+(\frac{2x}{y})\times y'=4y^{3}y'+1\)

\(3y'+\frac{2xy'}{y}-4y^{3}y'=1-\ln y^{2}\) (collect like terms, logarithm power law)

\(yy'3+\frac{2x}{y}-4y^{3}=1-\ln y^{2}\) (factor)

\(\displaystyle{y}'{\left(\frac{{{3}{y}+{2}{x}-{4}{y}^{{4}}}}{{y}}\right)}={1}-{{\ln{{\left({y}\right)}}}^{{2}}}\) (common divisor for the fraction)

\(\displaystyle{y}'={\frac{{{y}-{y}{\ln{{y}}}^{{{2}}}}}{{{3}{y}+{2}{x}-{4}{y}^{{{4}}}}}}\) Answer

(Try symbolab for computational questions like this)