If $y=e^{x^2-3x}$. Then value of $D_{x}y$ is
What i try::
$$D_{x}=\frac{\partial y}{\partial x}=e^{x^2-3x}\cdot (2x-3)$$
$$D_{x}y=\frac{\partial }{\partial y}\bigg(\frac{\partial y}{\partial x}\bigg)=\frac{\partial }{\partial y}\bigg(e^{x^2-3x}\cdot (2x-3)\bigg)=0$$
Is my solution is right. If not , Then How do i solve it . Help me please. Thanks