Answer
$y=-\dfrac{1}{4}x+\dfrac{1}{2}$
Work Step by Step
We have: $y(x)=(x^2-3x)^{-2}; x=1$
Now, $y(1)=[(1)^2-3(1)^{-2}]=\dfrac{1}{4}$
We differentiate both sides with respect to $x$.
$y^{\prime}(x)=-2 (x^2-3x)^{-3}(2x-3)$
The equation of a tangent line at $(1, \dfrac{1}{4})$ is:
$y-y_1=m(x-x_1)\\ y-\dfrac{1}{4}=-\dfrac{1}{4}x+\dfrac{1}{4}\\ y=-\dfrac{1}{4}x+\dfrac{1}{2}$
