Ok, the problem isn't hard if you know what those partials mean, it's just a huge mess haha To take the partial with respect to $t$ (or any variable) you just treat the other variables like constants. For example, if $P = xt^2$ then $\partial P/\partial t = 2xt$. So to compute the partial $\partial P / \partial t$ first use the product rule and take the derivative of the left summand:
$$\partial P / \partial t = \frac{\partial}{\partial t}\left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right) + \left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \frac{\partial}{\partial t} \left( \exp\big(-\frac{(x-\mu t)^2}{2\sigma^2t}\big) \right) =$$
$$= \left( \frac{-1}{t^2 \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right) + \left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \frac{\partial}{\partial t} \left( \exp\big(-\frac{(x-\mu t)^2}{2\sigma^2t}\big) \right)$$
Now to take the derivative on the right, we use the chain rule for $(e^{f(t)})' = f'(t)e^{f(t)}$:
$$\frac{\partial}{\partial t} \left( \exp\big(-\frac{(x-\mu t)^2}{2\sigma^2t}\big) \right) = \frac{\partial}{\partial t} \left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t} \right) =$$
$$ = - \left( \frac{-4\mu \sigma^2 t(x-\mu t) - 2\sigma^2(x-\mu t)^2}{4 \sigma^4 t^2}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t} \right) =$$
$$ = \frac{2\mu t(x-\mu t)+(x-\mu t)^2}{2 \sigma^2 t^2}\cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t} \right)$$
Where the first equality is just the well-known quotient rule for derivatives, and the second equality is just distributing the minus sign and reducing. I'd double-check these calculations if I were you, I just woke up. Putting these together and writing $A = \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right)$, we get the atrocious:
$$\partial P/\partial t = \left( \frac{-1}{t^2 \cdot \sqrt{2\pi\sigma^2}}\right) \cdot A + \left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \frac{2\mu t(x-\mu t)+(x-\mu t)^2}{2 \sigma^2 t^2} \cdot A =$$
$$ =\frac{A}{t^2 \cdot \sqrt{2 \pi \sigma^2}} \cdot \left( \frac{2\mu t(x-\mu t)+(x-\mu t)^2}{2 \sigma^2 t} - 1\right) =$$
$$= \frac{A}{t^2 \cdot \sqrt{2 \pi \sigma^2}} \cdot \left( \frac{x^2 - \mu^2t^2}{2 \sigma^2 t} - 1\right)$$
Now you need to check that the functional equation is satisfied, taking the partials wrt $x$. Hopefully it just falls right out - which it basically always does when something like that is true. But without some heavy theorems I see no way to prove this without doing the monster calculations. The partial wrt $x$ is a lot simpler, thankfully.
