Given $f(z) = (x^2+y)+i(xy)$ and
we integrate it using the Parabola Contour.
For a parabola, $\gamma(t) = t + it^2$.
So, $f(\gamma(t)) = 2t^2 + it^3$. What was done here is all the $y$ were replaced with $t^2$. Now, my understanding is this:
t + i(t^2)
x + i(y)
So, we replace all the $y$ in $f(x)$ with whatever corresponds to $y$ in $\gamma(t)$.
Is my understanding of $f(\gamma(t))$ correct ?
If $z = x+iy$, then the imaginary part of $z$ is $y$, and the real part is $x$.
When we take the curve $\gamma$, parameterized by $t$, we get a set of points in the complex plane. These points have a real and imaginary part.
Since $\gamma(t) = t+it^2$, and $t$ is real, the imaginary part of any point on $\gamma$ is $t^2$, and the real part is $t$.
When you set $z = \gamma(t)$ and plug this into $f(z)$, you have that $f(\gamma(t)) = \mathrm{Re}(\gamma(t))^2+\mathrm{Im}(\gamma(t)) + i\left(\mathrm{Re}(\gamma(t))\mathrm{Im}(\gamma(t))\right)$, which gets you to the result you've found.