Answer
$d=\sqrt{5x^2-16x+13}$
$(3,6)$; $\left(\dfrac{1}{5},\dfrac{2}{5}\right)$
Work Step by Step
Because the point $(x,y)$ is on the line $y=2x$, the coordinates can be written as $(x,2x)$.
The distance $d$ between $(x,2x)$ and $(2,3)$ can be written as:
$$\begin{align*}
d&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\
&=\sqrt{(x-2)^2+(2x-3)^2}\\
&=\sqrt{x^2-4x+4+4x^2-12x+9}\\
&=\sqrt{5x^2-16x+13}.
\end{align*}$$
We solve the equation $d=\sqrt{10}$ for $x$:
$$\begin{align*}
\sqrt{5x^2-16x+13}&=\sqrt{10}\\
5x^2-16x+13&=10\\
5x^2-16x+3&=0\\
(5x^2-15x)-(x-3)&=0\\
(x-3)(5x-1)&=0\\
x-3=0&\text{ or }5x-1=0\\
x=3&\text{ or }x=\dfrac{1}{5}\\
\end{align*}$$
The two points are:
$$(3,6)\text{ and }\left(\dfrac{1}{5},\dfrac{2}{5}\right).$$
