If $x^e=e^{x^2+y^2}$, then find $\frac{dy}{dx}$.

Given: $x^e=e^{x^2+y^2}$ Taking the natural logarithm on both sides:
$$\ln (x^e)=\ln (e^{x^2+y^2})$$
$$e\ln x=x^2+y^2$$
Differentiate both sides with respect to $x$:
$$\frac{d}{dx}(e\ln x)=\frac{d}{dx}(x^2+y^2)$$
$$\frac{e}{x}=2x+2y\frac{dy}{dx}$$
Rearrange to solve for $\frac{dy}{dx}$:
$$2y\frac{dy}{dx}=\frac{e}{x}-2x$$
$$2y\frac{dy}{dx}=\frac{e-2x^2}{x}$$
$$\frac{dy}{dx}=\frac{e-2x^2}{2xy}$$