Answer:
[tex]\frac{dy}{dx}=\frac{ye^{xy}+2x}{2y-xe^{xy}}[/tex]
Step-by-step explanation:
[tex]\frac{d}{dx} (e^{xy}+x^2-y^2=10)\\ye^{xy}+xe^{xy}\frac{dy}{dx} +2x-2y\frac{dy}{dx}=0\\\frac{dy}{dx}(2y-xe^{xy})=ye^{xy}+2x\\\frac{dy}{dx}=\frac{ye^{xy}+2x}{2y-xe^{xy}}[/tex]
