If x = b sec3θ and y = a tan3θ, prove that

$x=b \sec ^{3} \theta$

$\Rightarrow \sec ^{3} \theta=\frac{x}{b}$

$\Rightarrow \sec \theta=\left(\frac{x}{b}\right)^{\frac{1}{3}} \quad \ldots \ldots(1)$

Also,

$y=a \tan ^{3} \theta$

$\Rightarrow \tan ^{3} \theta=\frac{y}{a}$

$\Rightarrow \tan \theta=\left(\frac{y}{a}\right)^{\frac{1}{3}} \quad \ldots \ldots(2)$

We know

$\sec ^{2} \theta-\tan ^{2} \theta=1$

$\Rightarrow\left[\left(\frac{x}{b}\right)^{\frac{1}{3}}\right]^{2}-\left[\left(\frac{y}{a}\right)^{\frac{1}{3}}\right]^{2}=1 \quad[$ Using $(1)$ and $(2)]$

$\Rightarrow\left(\frac{x}{b}\right)^{\frac{2}{3}}-\left(\frac{y}{a}\right)^{\frac{2}{3}}=1$