Let $f(x, y)=x^{2}-7 x y, \mathbf{r}(t)=\langle\cos (t), \sin (t)\rangle$.
Use the Chain Rule to calculate $\frac{d}{d t} f(\mathbf{r}(t))$ at the value $t=(0)$.
(Use symbolic notation and fractions where needed.)

Step 1 ：Given the functions \(f(x, y) = x^2 - 7xy\) and \(r(t) = <\cos(t), \sin(t)>\).

Step 2 ：Compute the partial derivatives of \(f\), \(\frac{\partial f}{\partial x} = 2x - 7y\) and \(\frac{\partial f}{\partial y} = -7x\).

Step 3 ：Compute the derivatives of \(r\), \(\frac{dr}{dt} = <-\sin(t), \cos(t)>\).

Step 4 ：Apply the chain rule, \(\frac{df}{dt} = \frac{\partial f}{\partial x} * \frac{dx}{dt} + \frac{\partial f}{\partial y} * \frac{dy}{dt}\). Substituting the values we found, we get \(\frac{df}{dt} = (2x - 7y)(-\sin(t)) + (-7x)(\cos(t))\).

Step 5 ：Substitute \(r(t)\) into the equation, \(x = \cos(t)\) and \(y = \sin(t)\), so we substitute these values into the equation: \(\frac{df}{dt} = (2\cos(t) - 7\sin(t))(-\sin(t)) - 7\cos(t)(\cos(t))\). Simplify this to get \(\frac{df}{dt} = -2\cos(t)\sin(t) + 7\sin^2(t) - 7\cos^2(t)\).

Step 6 ：Evaluate at \(t = 0\), substitute \(t = 0\) into the equation to find the value of \(\frac{df}{dt}\) at \(t = 0\): \(\frac{df}{dt} = -2\cos(0)\sin(0) + 7\sin^2(0) - 7\cos^2(0)\). Simplify this to get \(\frac{df}{dt} = -7\).

Step 7 ：So, the derivative of \(f(r(t))\) with respect to \(t\) at \(t = 0\) is \(\boxed{-7}\). This is the final answer.