Answer
Using the fact that ${\bf{r}}'\left( t \right) = k\left( t \right)\nabla {f_{{\bf{r}}\left( t \right)}}$ for some positive scalar function $k\left( t \right)$, we show that:
$\frac{{y'\left( t \right)}}{{x'\left( t \right)}} = \frac{{{f_y}}}{{{f_x}}}$
Work Step by Step
If the path ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$ follows the gradient of $f\left( {x,y} \right)$, then the tangent vector ${\bf{r}}'\left( t \right)$ points in the direction of $\nabla f$ for all $t$. That is, ${\bf{r}}'\left( t \right) = k\left( t \right)\nabla {f_{{\bf{r}}\left( t \right)}}$ for some positive scalar function $k\left( t \right)$. Since ${\bf{r}}'\left( t \right) = \left( {x'\left( t \right),y'\left( t \right)} \right)$ and $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( {{f_x},{f_y}} \right)$, so
$\left( {x'\left( t \right),y'\left( t \right)} \right) = k\left( t \right)\left( {{f_x},{f_y}} \right)$
In component forms, we get
$x'\left( t \right) = k\left( t \right){f_x}$ ${\ \ }$ and ${\ \ }$ $y'\left( t \right) = k\left( t \right){f_y}$
Thus, $k\left( t \right) = \frac{{x'\left( t \right)}}{{{f_x}}} = \frac{{y'\left( t \right)}}{{{f_y}}}$. Hence,
$\frac{{y'\left( t \right)}}{{x'\left( t \right)}} = \frac{{{f_y}}}{{{f_x}}}$
