If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$ If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$

It is very long to direct differentiate it.Can someone help me?
 A: $\dfrac{c^4}x+\dfrac{s^4}y=\dfrac 1{x+y}\iff(c^4y+s^4x)(x+y)=xy\iff s^4x^2+\underbrace{(c^4+s^4-1)}_{-2s^2c^2}xy+c^4y^2=0$
Thus $(s^2x-c^2y)^2=0\iff y=\tan(\alpha)^2x$
A: Using cauchy schwarz inequality $$\frac{(\cos^2\alpha)^2}{x}+\frac{(\sin^2 \alpha)^2}{y}\geq \frac{\cos^2 x+\sin^2 x}{x+y}=\frac{1}{x+y}$$.
And equality holds when $$\frac{\cos^2 \alpha}{x}=\frac{\sin^2 x}{y}$$
So $$y=\tan^2 (\alpha)\cdot x\Rightarrow \frac{dy}{dx}=\tan^2(\alpha)$$.
A: Denote $a=\sin \alpha$ and $b=\cos \alpha$. Then:
$$xy=(x+y)(xa^4+yb^4) \qquad (1) \Rightarrow \\
xy=x^2a^4+y^2b^4+xy(b^4+a^4) \Rightarrow \\
xy=x^2a^2(1-b^2)+y^2b^2(1-a^2)+xy(1-2a^2b^2) \overbrace{\Rightarrow}^{:a^2b^2} \\
(x+y)^2-\frac{x^2}{b^2}-\frac{y^2}{a^2}=0 \Rightarrow F(x,y)=0.$$
Take the derivative:
$$y'=-\frac{F_x}{F_y}=-\frac{2(x+y)-\frac{2x}{b^2}}{2(x+y)-\frac{2y}{a^2}}\overbrace{=}^{(1)}-\frac{\frac{xy}{xa^4+yb^4}-\frac{x}{b^2}}{\frac{xy}{xa^4+yb^4}-\frac{y}{a^2}}=\\
-\frac{xyb^2-x^2a^4-xyb^4}{xya^2-xya^4-y^2b^4}\cdot \frac{a^2}{b^2}=\frac{a^2}{b^2},$$
because:
$$-\frac{xyb^2-x^2a^4-xyb^4}{xya^2-xya^4-y^2b^4}=1 \iff \\
xy=(x+y)(xa^4+yb^4) \qquad (1)$$
