I'm a bit rusty on epsilon delta proofs so I wanted to double check that my proof is correct:
Let $\epsilon > 0$ and $\delta = \epsilon * \frac{\sqrt{2}}{3}$. Then for $(x,y)$ and $(u,v)$, $\sqrt{(x-u)^2 + (y-v)^2} < \delta$, we have $(x-u)^2 + (y-v)^2 < \epsilon ^2 * \frac{2}{9}$, so $(x-u)^2,(y-v)^2 < \epsilon ^2 * \frac{2}{9}. $ This means $2(x-u) < \frac{2*\sqrt{2}}{3}*\epsilon$ and $3(y-v) < \epsilon * \sqrt{2}$. Finally, we have:
$$|2(x-u) - 3(y-v)| = |(2x-3y) - (2u-3v)| < |\frac{2*\sqrt{2}}{3}*\epsilon - \sqrt{2}*\epsilon| = \frac{\sqrt{2}}{3}*\epsilon < \epsilon.$$
I know what to show but I think I got an uglier delta than needed. That or I messed up trying to show the final inequality.
Thank you!
Your proof is absolutely correct. But, if you want to find a "nicer" delta for given epsilon, I would suggest you to use the following strategy.
First, take a look at $\left| f \left( x, y \right) - f \left( u, v \right) \right|$. Simplify as much as possible. To simplify it, try getting all the terms in $x, u$ and $y, v$ in terms of $\sqrt{\left( x - u \right)^2 + \left( y - v \right)^2}$.
For doing so, observe that $\left| x - u \right| \leq \sqrt{\left( x - u \right)^2 + \left( y - v \right)^2}$ and $\left| y - v \right| \leq \sqrt{\left( x - u \right)^2 + \left( y - v \right)^2}$.
Therefore, we now have
\begin{align*} \left| f \left( x, y \right) - f \left( u, v \right) \right| &= \left| 2x - 3y - 2u + 3v \right| \\ &= \left| 2 \left( x - u \right) + 3 \left( v - y \right) \right| \\ &\leq 2 \left| x - u \right| + 3 \left| y - v \right| \\ &\leq 5 \sqrt{\left( x - u \right)^2 + \left( y - v \right)^2} \end{align*}
Now, we have to find a $\delta$ such that $\sqrt{\left( x - u \right)^2 + \left( y - v \right)^2} < \delta$ would imply that $\left| f \left( x, y \right) - f \left( u, v \right) \right| < \epsilon$. From above, you can see that if we choose $\delta = \dfrac{\epsilon}{5}$, our job is done.
Now, we are ready to write a "textbook proof" as you have written in your attempt!