The point $P$ of the curve $y^2=2x^3$ such that the tangent at $P$is perpendicular to the line $4x-3y+2=0$ is given by

Let $P\left(x_1,y_1\right)$ be a point on $y^2=2x^3$ such that the

tangent at $P$ is perpendicular to the line $4x-3y+2=0$.

$\therefore {\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\times \left(\frac{-4}{-3}\right)=-1\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left\{x_1,y_1\right\}}=\frac{3}{4}$               ..(i)

Now, 

$\begin{array}{l}{y}^2=2x^3\\ \Rightarrow 2y\frac{dy}{dx}=6x^2\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{3{x_1}^2}{y_1}\end{array}$                             …(ii)

Frum (i) and (ii), we get

$y^2=2x^3$$\frac{3x_1^2}{y_1}-\frac{3}{4}\Rightarrow y_1-4x_1^2$                                              …(iii)

Since $\left(x_1,y_1\right)$ lies on $y^2=2x^3$

$\therefore {y_1}^2=2{x_1}^3$                                                                  …(iv)

Solving (iii) and (iv), we get

${\left(4x_1\right)}^2=2x_1^3\Rightarrow x_1=0,x_1=1/8$ 

Putting the values of $x_1$ in (iv), we get

$y_1=0,y_1=\pm \frac{1}{16}$

Hence, the required points are $(0, 0), (1/8, 1/16) ,(1/8,-1 /16)$