We have,
$\begin{aligned}
& \cos ^{-1}\left(\frac{x}{2}\right)+\cos ^{-1}\left(\frac{y}{3}\right)=\theta \\
& \Rightarrow \cos ^{-1}\left[\frac{x}{2} \cdot \frac{y}{3}-\sqrt{\left(1-\left(\frac{x}{2}\right)^2\right)\left(1-\left(\frac{y}{3}\right)^2\right)}\right]=\theta \\
& \Rightarrow \frac{x}{2} \cdot \frac{y}{3}-\sqrt{\left(1-\left(\frac{x}{2}\right)^2\right)\left(1-\left(\frac{y}{3}\right)^2\right)}=\cos \theta \\
& \quad=\frac{x y}{6}-\cos \theta=\sqrt{\left(1-\left(\frac{x}{2}\right)^2\right)\left(1-\left(\frac{y}{3}\right)^2\right)}
\end{aligned}$
squaring both sides, we get
$\begin{aligned}
& \Rightarrow\left(\frac{x y}{6}-\cos \theta\right)^2=\left(1-\frac{x^2}{4}\right)\left(1-\frac{y^2}{9}\right) \\
& \begin{aligned}
\Rightarrow \frac{x^2 y^2-12 x y \cos \theta+36 \cos ^2 x}{36}=\frac{\left(4-x^2\right)\left(9-y^2\right)}{36} & =36-9 x^2-4 y^2+x^2 y^2 \\
\Rightarrow x^2 y^2-12 x y \cos \theta+36 \cos ^2 x & =36\left(1-\cos ^2 \theta\right)
\end{aligned} \\
& \Rightarrow 9 x^2-12 x y \cos \theta+4 y^2=36 \sin ^2 \theta
\end{aligned}$