The equation of the tangent to the curve $x=2{\cos }^3\theta$ and $y=3{\sin }^3\theta$ at the point $\theta =\pi /4$ is :

${\left.x\right|}_{\theta =\frac{\pi }{4}}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$

${\left.y\right|}_{\theta =\frac{\pi }{4}}=\frac{3}{2\sqrt{2}}$

${\left.{\left.\frac{dy}{dx}\right|}_{\theta =\frac{\pi }{4}}\frac{9{\sin }^2\theta \cos \theta }{-6{\cos }^2\theta \sin \theta }\right|}_{\theta =\frac{\pi }{4}}=\frac{-3}{2}$

$\therefore$ Equation of tangent is $\left(y-\frac{3}{2\sqrt{2}}\right)=\frac{-3}{2}\left(x-\frac{1}{\sqrt{2}}\right)$

$\begin{array}{l}\Rightarrow 3\sqrt{2}x+2\sqrt{2}y=6\\ \Rightarrow 3x+2y=3\sqrt{2}.\end{array}$