If $\cos \left({\mathrm{x}}^2{\mathrm{y}}^3\right)+\sin \left({\mathrm{x}}^2{\mathrm{y}}^3\right)=\mathrm{k}$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$

The given equation is $\cos \left({\mathrm{x}}^2{\mathrm{y}}^3\right)+\sin \left({\mathrm{x}}^2{\mathrm{y}}^3\right)=\mathrm{k}$

Use the implicit differentiation, differentiate both sides with respect to $x$

$\left[-\sin \left(x^2{y}^3\right)+\cos \left(x^2{y}^3\right)\right]\left[2x{y}^3+3x^2{y}^2\frac{dy}{dx}\right]=0$

It implies that 

$\begin{array}{rcl}2x{y}^3+3x^2{y}^2\frac{dy}{dx}& =& 0\\ 2y+3x\frac{dy}{dx}& =& 0\\ \frac{dy}{dx}& =& \boxed{\mathbf{-}\frac{\mathbf{2}\mathbf{y}}{\mathbf{3}\mathbf{x}}}\end{array}$