Let $P(x_0,y_0)$ be the point on the hyperbola $3x^2-4y^2=36,$ which is nearest to the line $2y+3x=1.$ Then $\sqrt{2}(y_0+x_0)$ is equal to:

$3x^2-4y^2=36$

$\frac{dy}{dx}=\frac{3x}{4y}=\frac{3x_0}{4y_0}$

Point $(x_0,y_0)$ on curve.

$\frac{3x_0}{4y_0}=\frac{-3}{2}$

$x_0=-2y_0$

$3x_0^2-4y_0^2=36$

$y_0=\pm 3/\sqrt{2}, x_0=-2y_0$

$(\frac{-6}{\sqrt{2}},\frac{3}{\sqrt{2}}),(\frac{6}{\sqrt{2}},\frac{-3}{\sqrt{2}})$

Calculating $\perp$ distance from both co-ordinate $(\frac{6}{\sqrt{2}},\frac{-3}{\sqrt{2}})$ is nearest to line.