If you are integrating $$\int_\gamma y^2\,dz$$ Where $\gamma$ is the line segment from $1$ to $i$. You parameterize the line $$x(t)=1-t$$ $$y(t)=t$$ $$\implies z(t)=1-t+it$$
Now, if you want to use the formula: $$\int_\gamma f(z(t))z'(t)\,dt,$$ would you have the integral $$\int_0^1(t)^2(-1+i)\,dt$$ or would you have $$\int_0^1(it)^2(-1+i)\,dt.$$ I'm assuming its the first integral because you want the imaginary part, which is just $t$ and not $it$.
It's the first one: $\int_0^1 t^2(-1 + i)\, dt$. The choice of $f(z)$ is $f(z) = \text{Im}(z)^2$. So with $z(t) = (1 - t) + it$, we have $z'(t) = -1 + i$ and thus
$$\int_\gamma y^2\, dz = \int_0^1 f(z(t))z'(t)\, dt = \int_0^1 t^2(-1 + i)\, dt = \frac{-1 + i}{3}.$$