Let $\varphi: M \to N$ a smooth map between manifolds. Given $p\in M$, we can consider the differential of $\varphi$ at $p$, which is a linear map $d\varphi_p: T_pM \to T_{\varphi(p)}N$ that brings the points of the tangent space of $M$ at $p$ to points of the tangent space of $N$ at $\varphi(p)$.
Assume $M = \{(x,y): y = x^4-3x^2\}$ and $N = \{(x,y): y = x^2\}$, and consider $\varphi:M\to N$ given ny $\varphi(x,y) = (x, x^2)$. Let $p = (\sqrt{3/2}, -9/4)$. It is clear that $\varphi(p) = (\sqrt{3/2}, 3/2)$ and it is easy to check that $$ d\varphi_p = \begin{pmatrix} 1 & 0 \\ \sqrt{6} & 0 \end{pmatrix}. $$
The tangent line to M at p is $y = -9/4$ (it is clear from the graph of the function) and the tangent line to N at $\varphi(p)$ is $y = \sqrt{6}x-3/2$ (the slope is the derivative of $f(x) = x^2$ at $\sqrt{3/2}$ and the tangent line must cross $\varphi(p) = (\sqrt{3/2}, 3/2)$).
According to the properties of the differential (pushforward), $d\varphi_p$ should take points of the straight line $y = -9/4$ to points of the straight line $y = \sqrt{6}x-3/2$. To find $d\varphi_p(T_pM)$, we compute $$ \begin{pmatrix} 1 & 0 \\ \sqrt{6} & 0 \end{pmatrix}\begin{pmatrix} x\\ -9/4 \end{pmatrix} =\begin{pmatrix} x\\ \sqrt{6}x \end{pmatrix} $$
These points, $(x, \sqrt{6})$, should satisfy the equation $y = \sqrt{6}x-3/2$, but they don't. My question is, where is the problem? I checked the calculations, did I misapply the theory?