Problem:
Given the smooth function $f(x,y)$ such as $f_{x}(1,1)=2$, and $f_{y}(1,1)=-3$, and $\vec{r}(t)=(t,t^{2})$, which is a parabola parameterization. Then the derivative of the composite function $g(t)=(f\circ \vec{r})(t)=f(\vec{r}(t))$ at $t=1$ is equal to $1$. [True or False]
Solution:
$g(t)=(f\circ\vec{r})\vec{r}\ '$
$=f\ '(\vec{r}(t))(1,2t)$
$=\begin{equation} \left.f\ '((t,t^{2}))(1,2t)\right\vert_{t=1,}^{}=f\ '(1,1)\cdot(1,2)=(2,-3)\cdot(1,2)=-4\neq1 \end{equation}$
Hence, the statement is false.
Is that correct?
I don't like your notation. I'll go with the standard chain rule: $$\frac{d}{dt}\big|_{t=1}g(t)=\frac{d}{dt}\big|_{t=1}f(r(t))=\frac{df}{dx}(r(t))\frac{dr_1}{dt}(t)+\frac{df}{dy}(r(t))\frac{dr_2}{dt}(t)\big|_{t=1}$$ as $r_1(t)=t$ and $r_2(t)=t^2$ we obtain $\frac{dr_1}{dt}=1,\frac{dr_2}{dt}=2t$. Evaluating the whole at $t=1$ $$\frac{d}{dt}\big|_{t=1}g(t)=f_x(1,1)\cdot1+f_y(1,1)2\cdot 1=-4.$$ So yes, your result is correct.