Let $f: \Bbb{R}^2 \rightarrow \Bbb{R}^3$ be defined via
$$(s,t) \mapsto (\sin t, st^2, s^2-1)$$
Let $\omega$ be the $1$-form
$$\omega=dx + x \, dy+y^2 \, dz$$
I computed $f^*\omega$ to being (the pull back under $\omega$)
$$\cos t + 2t \sin t \, dt \,+2s^3t^6 \, ds$$
Did I compute it correctly? I dont think so, so I recomputed below:
I did
\begin{align} f^*\omega&=f^*(dx+x \, dy+y^2\, dz)\\ &=d(\sin t)+\sin t \, d(st^2) + s^2t^4 \,d(s^2-1)\\ &=\cos t dt + \sin t(t^2 \,ds +2st\, dt) +2s^3t^4 \, ds\\ &= \cos t \,dt + t^2 \sin t \,ds + 2st \sin t \,dt +2s^3t^4 \,ds \\ &=(\cos t + 2st \sin t) \, dt + (t^2 \sin t + 2s^3t^4)\, ds \end{align}
is THIS correct^?