In case we have a dot product of $v=(v^1,v^2,v^3)$ and $u=(u^1,u^2,u^3)$ we write the dot product in Einstein notation as follow:
$$\delta_{i j}u^iv^j$$
Is the following correct?
To "open" this we only look when $i=j$ as $\delta_{ij}$ is the Kronecker delta so we have:
$\delta_{11}u^{1}v^1+\delta_{22}u^{2}v^2+\delta_{33}u^{3}v^3=\delta_{1}uv^1+\delta_{2}u^{}v^2+\delta_{3}uv^3=\delta_{1}v^1u+\delta_{2}v^2u+\delta_{3}v^3u=\delta vu+\delta vu+\delta v u$
Your working is not correct. I don't know how $\delta_{11}u^1v^1$ became $\delta_1uv^1$ and then became $\delta uv$. Neither of these last two expressions make sense.
Recall that the Kronecker delta function is defined by
$$\delta_{ij} = \begin{cases} 1 & i = j\\ 0 & i \neq j. \end{cases}$$
Therefore, $\delta_{ij}u^iv^j$ reduces to $\delta_{11}u^1v^1 + \delta_{22}u^2v^2 + \delta_{33}u^3v^3 = u^1v^1 + u^2v^2 + u^3v^3$ which is the usual expression for the dot product of $(u^1, u^2, u^3)$ and $(v^1, v^2, v^3)$.