Let the distribution $\displaystyle p(\theta | x) = L(\theta | x)p(\theta) = p_{1} = \exp[-\frac{1}{2} ((\frac{\theta - x}{\sigma})^{2} + (\frac{\theta - \mu_{0}}{\sigma_{0}})^{2})]$ where $L$ is the likelihood function for $\theta$
How can I get to distribution $p(\theta | x) = p(\theta) = p_2 = exp[\frac{-1}{2} \frac{(\theta - \mu_{1})^{2}}{\sigma_{1}^{2}}]$ via completing the square
with the help of $\displaystyle\frac{1}{\sigma_{1}^{2}} = \frac{1}{\sigma} + \frac{1}{\sigma_{0}^{2}}$, $\displaystyle\frac{\mu_{1}}{\sigma_{1}^{2}} = \frac{x}{\sigma^{2}} + \frac{\mu_{0}}{\sigma_{0}^{2}}$
Terribly rusty wit these. Any help is appreciated. Hints are encouraged.
Edit: corrected typos
I still think there is an error in your formula $$ \frac{1}{\sigma^2_1} = \frac{1}{\sigma^\color{red}{2}} + \frac{1}{\sigma_0^2}. $$
Assuming this is indeed an error, since we are given the formulae of $\sigma_1$, $\mu_1$, expanding gives $$ \frac{1}{\sigma_1^2} = \frac{1}{\sigma^2} + \frac{1}{\sigma^2_0} = \frac{\sigma^2 + \sigma_0^2}{\sigma^2 \sigma^2_0} \tag{1} $$ and $$ \frac{\mu_1}{\sigma_1^2} = \frac{x}{\sigma^2} + \frac{\mu_0}{\sigma^2_0} = \frac{\mu_0 \sigma^2 + x\sigma_0^2}{\sigma^2 \sigma^2_0} \tag{2}. $$
The trick here is that by assuming the new distribution is in $\theta$, we can discard any term not $\theta$-related and treat them as normalizaing constants. So $$ \begin{align*} \left[ \frac{(\theta - x)^2}{\sigma^2} \right] + \left[ \frac{(\theta - \mu_0)^2}{\sigma_{0}^2} \right] &= \frac{\sigma^2_0 (\theta^2 - 2x\theta + x^2) + \sigma^2 (\theta^2 - 2\mu_0 \theta + \mu_0^2)}{\sigma^2 \sigma_0^2} \\ &= \frac{\sigma^2 + \sigma^2 _0}{\sigma^2 \sigma_0^2} \theta^2 - 2\theta \frac{x \sigma_0^2 + \mu_0 \sigma^2}{\sigma^2 \sigma_0^2} + C \\ &= \frac{\theta^2}{\sigma_1^2}- 2 \theta \frac{\mu_1}{\sigma_1^2} + C \tag{From (1) and (2)} \\ &= \frac{\theta^2}{\sigma_1^2} - 2 \theta \frac{\mu_1}{\sigma_1^2} + \color{blue}{\frac{\mu_1^2}{\sigma_1^2}} - \color{blue}{\frac{\mu_1^2}{\sigma_1^2}} + C \\ &= \frac{(\theta - \mu_1)^2}{\sigma_1^2} + C', \end{align*} $$ where $C$, $C'$ are normalizing constants not depending on $\theta$. This gives you the desired Gaussian form.