A curve is defined by the equation $x^2+y^2=16(x^2-y^2)$. Find all points on the curve at which the tangent line is horizontal.
I have differentiated the equation and found $\frac {dy}{dx} = \frac {8x-xy^2-x^3}{x^2y+y^3+8y}.$ Then, I don't know how to solve $8x-xy^2-x^3 = 0.$
Also, I computed the limit shown below. The answer is 2, but the correct answer is $\ln 12^\frac13$:
$\lim_{x\to0^+} \left(\frac133^x+\frac232^x\right)^\frac1x$
$=\lim_{x\to0^+} \left(\frac{\frac133^x\ln3+\frac232^x\ln2}{(\frac133^x+\frac232^x)}\right)$ $=\lim_{(\frac{2}{3})^x\to0^+} \left(\frac{\frac13\ln3+\frac23\frac23^x\ln2}{(\frac13+\frac23(\frac23)^x)}\right)$
How can I solve it?