The coordinates of the point P on the curve ${\mathrm{y}}^2=2{\mathrm{x}}^3$, 

the tangents at which is perpendicular to the line

 $4\mathrm{x}-3\mathrm{y}+2=0$, are given by

Differentiating ${\mathrm{y}}^2=2{\mathrm{x}}^3$, we get
$2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=6{\mathrm{x}}^2\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3{\mathrm{x}}^2}{\mathrm{y}}$
The slope of the line $4\mathrm{x}-3\mathrm{y}+2=0\text{ is }\frac{4}{3}$. Therefore, the coordinates of $\mathrm{P}(\mathrm{x},\mathrm{y})$must
satisfy
$\frac{3{\mathrm{x}}^2}{\mathrm{y}}\cdot \frac{4}{3}=-1\Rightarrow 4{\mathrm{x}}^2=-\mathrm{y}$
Also, ${\mathrm{y}}^2=2{\mathrm{x}}^3$. Solving these, we get $\mathrm{x}=\frac{1}{8}$ and $\mathrm{y}=-\frac{1}{16}$(clearly $\mathrm{x}\ne 0)$