Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 352: 64

$\frac{dy}{dx}=\frac{-2x-ye^{xy}}{xe^{xy}-2y}$

$e^{xy}+x^2-y^2=10$ $((1)y+x\frac{dy}{dx})e^{xy}+2x-2y\frac{dy}{dx}=0$ $ye^{xy}+(xe^{xy}\frac{dy}{dx})+2x-(2y\frac{dy}{dx})=0$ $(xe^{xy}-2y)\frac{dy}{dx}=-2x-ye^{xy}$ $\frac{dy}{dx}=\frac{-2x-ye^{xy}}{xe^{xy}-2y}$