Specific example of $\mathrm{d}(F^* \alpha) = F^* (\mathrm{d} \alpha)$

Question

I have to prove the equality for this specific example and I am lost somewhere close to the end.

Let $F: \mathbb{R}^2 \to \mathbb{R}^3 : F(x^1, x^2) = (x^1- x^2, x^1 + x^2, (x^1)^2)$ and $\alpha$ be the 2-form such that $\alpha(y) = (y^1)^2 \mathrm{d}y^1 \wedge \mathrm{d}y^2$

For the left side of the equality:

$F^* \alpha = (y^1)^2 \circ F(x) \mathrm{d}(y^1 \circ F(x)) \wedge \mathrm{d}(y^2 \circ F(x))$

$= (x^1- x^2)^2 \mathrm{d}(x^1- x^2) \wedge \mathrm{d}(x^1 + x^2)$ $= (x^1 - x^2)^2 (\mathrm{d}x^1 - \mathrm{d}x^2) \wedge (\mathrm{d}x^1 + \mathrm{d}x^2)$

$= 2(x^1 - x^2)^2 \mathrm{d}x^1 \wedge \mathrm{d}x^2$

$\mathrm{d}(F^* \alpha) = \mathrm{d}[2(x^1-x^2)^2 \mathrm{d}x^1 \wedge \mathrm{d}x^2] = \mathrm{d}[2(x^1 - x^2)^2 \mathrm{d}x^1] \wedge \mathrm{d}x^2$

$= 4(x^1 - x^2)\mathrm{d}x^1 \wedge \mathrm{d}x^2$

My problem appears when calculating the right side

$\mathrm{d}\alpha = \mathrm{d}[(y^1)^2 \mathrm{d}y^1 \wedge \mathrm{d}y^2] = \mathrm{d}[(y^1)^2\mathrm{d}y^1]\wedge \mathrm{d}y^2$

Using the formula for 1-forms

$$\mathrm{d}A=\sum_{i < j} ^ n (\partial_{y^i}A^j - \partial_{y^j}A^i)\mathrm{d}y^i\wedge \mathrm{d}y^j$$

I get that

$\mathrm{d}\alpha = (\frac{\partial 0}{\partial y^1} - \frac{\partial (y^1)^2}{\partial y^2}) \mathrm{d}y^1 \wedge \mathrm{d}y^2 = 0$

Therefore I can't continue with the equality

Answer 1

Found the mistake (I think).

$\mathrm{d}(F^*\alpha) = \mathrm{d}[2(x^1-x^2)^2\mathrm{d}x1] \wedge \mathrm{d}x^2$

actually equals

$= 4(x^1-x^2)\mathrm{d}x^1 \wedge \mathrm{d} x^2 \wedge \mathrm{d}x^2 = 0$