## Thomas' Calculus 13th Edition

$\frac{2e^{2x}-cos(x+3y)}{3cos(x+3y)}$
We start with: $e^{2x}=\sin(x+3y)$ Now use implicit differentiation: $2e^{2x}=(1+3y') \cos(x+3y)$ And solve for $y'$: $1+3y'=\frac{2e^{2x}}{cos(x+3y)}$ $3y'=\frac{2e^{2x}}{cos (x+3y)}-1$ $y'=\frac{2e^{2x}-cos(x+3y)}{3cos(x+3y)}$