Can someone confirm if my solution is right or if I have done something that is not permitted
$$ \begin{align} & \int_\gamma e^{\pi z}=\int_\gamma \left( \frac{ e^{\pi z}}{\pi}\right)' \, dz \\[8pt] = {} & \int_{-1}^1 \left(\frac{e^{\pi (3-3t^2+i(t^3-3t+1))}}{\pi}\right)' (3-3t^2+i(t^3-3t+1))' \,dz \\[8pt] ={} & \int_{-1}^1 \left(\frac{e^{\pi (3-3t^2+i(t^3-3t+1))}}{\pi} (3-3t^2+i(t^3-3t+1))\right)'dz \\[8pt] = {} & \frac{e^{\pi(3-3t^2 +i(t^3-3t+1))}}{\pi} (3-3t^2+i(t^3-3t+1)) \end{align} $$
$(f(t)g(t))' \neq f'(t) g'(t)$ as you have written (moving from the first line to the second). Also, in moving from the second statement to the third, you have to change your variable of integration (but this could just be a typo rather than a lack of understanding)
Instead, write
$\int_\gamma e^{\pi z} \,dz = \int_{-1}^1 e^{\pi \gamma(t)} \gamma'(t) \, dt$
then use the chain-rule (i.e. $u$-substitution) to evaluate: let $s = \gamma(t)$, then $ds = \gamma'(t) dt$, and you have
$$ \int_{\gamma(-1)}^{\gamma(1)} e^{\pi s} \,ds = \left.\dfrac{1}{\pi} e^{\pi s} \right|_{s = \gamma(-1)}^{\gamma(1)}. $$
Note that most of this can be bypassed by observing that $e^{\pi z}$ is analytic, so you can evaluate associated contour integrals by just taking the antiderivative ($\frac{1}{\pi} e^{\pi z}$) and evaluating at the endpoints of the contour (and in the special case the contour is closed, the result is zero).