The line v is vertical and intersects the x-axis at $\frac{4}{3}$.

Let's mark the required points with P(x,y)

Since line v is vertical, the distance of point P from line v is the horizontal distance. This distance determines the point V.

The x coordinate of point V is always $\frac{4}{3}$

On the basis of the task, you can write:

$$\overline{PF}:\overline{PV}=3:2$$

Since we are actually looking for the distance of the points, it is best to square the expression.

$${\overline{PF}}^2:{\overline{PV}}^2=9:4$$

$${\overline{PF}}^2={\left(x-3\right)}^2+{\left(y-0\right)}^2$$

$${\overline{PF}}^2=x^2-2·x·3+9+y^2$$

$${\overline{PV}}^2={\left[x-\frac{4}{3}\right]}^2+\bcancel{{\left(y-y\right)}^2}$$

$${\overline{PV}}^2={\left(x-\frac{4}{3}\right)}^2$$

$${\overline{PV}}^2=x^2-2·x·\frac{4}{3}+{\left(\frac{4}{3}\right)}^2$$

$$9\left(x^2-\frac{8}{3}x+\frac{16}{9}\right)=4\left(x^2-6x+9+y^2\right)$$

$$9x^2\bcancel{-24x}+16=4x^2\bcancel{-24x}+36+4y^2$$

$$9x^2-4x^2-4y^2=36-16$$

$$5x^2-4y^2=20$$

$$5x^2-4y^2=20 |·\frac{1}{20}$$

$$\frac{5x^2}{20}-\frac{4y^2}{20}=1$$