Naïve approach to the pushforward and a mistake

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Suppose we need to find an image of a vector field $X=x \frac{\partial}{\partial x}$ under a diffeomorphism $\phi: \ \mathbb{R} \to \mathbb{R}_+^*$, $x \mapsto y=e^x$. I want to use the definition of a pushforward as a differential of a map, i.e. $$\phi_* X = \mathrm{d} \phi (X),$$ Then let's calculate formally $$\mathrm{d} \phi (X) = \frac{\partial y}{\partial x} \mathrm{d} x \left(x \frac{\partial}{\partial x} \right)= \frac{\partial y}{\partial x}x \ \mathrm{d} x \left( \frac{\partial}{\partial x} \right) =\frac{\partial y}{\partial x}x =e^x x = y \ln y. $$
Here I used the fact that $\mathrm{d} x^i (\partial_j)= \delta^i_j$. However, the pushforward of a vector field should be again a vector field, i.e. the correct answer is $$ y \ln y \ \color{red}{ \frac{\partial}{\partial y} } $$

Could someone please tell me where I have such an obvious mistake in my reasoning?