Use calculus?

$\begin{aligned} 3x^2+2y^2 & = 6x \\ 3x^2-6x+2y^2 & = 0 \\ 3(x^2-2x+1) + 2y^2 & = 3 \\ (x-1)^2 + \frac 23 y^2 & = 1 \end{aligned}$

Using the identity $\cos^2 \theta + \sin^2 \theta = 1$ , we have $\begin{cases} \cos \theta = x-1 & \implies x = \cos \theta + 1 \\ \sin \theta = \sqrt {\dfrac 23}y & \implies y = \sqrt {\dfrac 32} \sin \theta \end{cases}$

Therefore, we have:

$\begin{aligned} x^2+y^2 & = (\cos \theta + 1)^2 + \frac 32 \sin^2 \theta \\ & = \cos^2 \theta + 2\cos \theta + 1 + \frac 32 \sin^2 \theta \\ & = 2 + 2\cos \theta + \frac 12 \sin^2 \theta \\ & = 2 + 2\cos \theta + \frac 12(1 - \cos^2 \theta) \\ & = \frac 52 - \frac 12(\cos^2 \theta - 4\cos \theta + 4) + 2 \\ & = \frac 92 - \frac 12({\color{#3D99F6}\cos \theta} - 2)^2 & \small \color{#3D99F6} \text{Note that }x^2+y^2 \text{ is maximum, when }\cos \theta = 1 \\ & \le \boxed{4} \end{aligned}$