The equation $$\frac{(x\cos\,\theta-y\sin\,\theta)^2}{a^2}+\frac{(x\sin\,\theta+y\cos\,\theta)^2}{b^2}=1 \ \ \ \ \ \ \ \ (*)$$
has the following expression for the derivative:
$$\frac{\mathrm dy}{\mathrm dx}=-\frac{a^2 x\,\sin^2\theta+y(a-b)(a+b)\sin\,\theta \cos\,\theta+b^2 x\,\cos^2\theta}{a^2 y\,\cos^2\theta+x(a-b)(a+b)\sin\,\theta \cos\,\theta+b^2 y\,\sin^2\theta}$$
Separating the numerator and denominator of this expression, and equating any of the two to zero, and then solving as a simultaneous equation with the above equation$(*)$ gives:
$$x=\pm\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}$$ and $$y=\pm\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$$
How?