If $x^{\frac{3}{4}}+y^{\frac{3}{4}}=a^{\frac{3}{4}}$ (a is a constant) then $\frac{dy}{dx}$ is equal to :

$-\left(\frac{y}{x}\right)^{\frac{1}{4}}$

The correct answer is Option (3) → $-\left(\frac{y}{x}\right)^{\frac{1}{4}}$

$x^{\frac{3}{4}}+y^{\frac{3}{4}}=a^{\frac{3}{4}}$

Taking differentiation both sides,

$\frac{3}{4}x^{-\frac{1}{4}}+\frac{3}{4}y^{-\frac{1}{4}}\left(\frac{dy}{dx}\right)=0$

$\frac{dy}{dx}=-\left(\frac{y}{x}\right)^{\frac{1}{4}}$