已知 $x^2 y - \tan^{-1} y = \frac{3}{4}\pi$。用 $x$ 和 $y$ 表示 $\frac{\text{d}y}{\text{d}x}$。

1. Find an expression for $\frac{\text{d}y}{\text{d}x}$ using implicit differentiation (from the preceding part).

Differentiating $x^2 y - \tan^{-1} y = \frac{3}{4}\pi$ with respect to $x$:
$2x y + x^2 \frac{\text{d}y}{\text{d}x} - \frac{1}{1+y^2}\frac{\text{d}y}{\text{d}x} = 0$
$\frac{\text{d}y}{\text{d}x}\left(x^2 - \frac{1}{1+y^2}\right) = -2xy$
$\frac{\text{d}y}{\text{d}x} = \frac{-2xy(1+y^2)}{x^2(1+y^2) - 1}$

2. Find the corresponding value of $x$ when $y = 1$.

Substitute $y = 1$ into the original equation:
$x^2(1) - \tan^{-1}(1) = \frac{3}{4}\pi$
$x^2 - \frac{\pi}{4} = \frac{3}{4}\pi$
$x^2 = \pi$
Since $x > 0$, $x = \sqrt{\pi}$.

3. Calculate the exact value of $\frac{\text{d}y}{\text{d}x}$.

Substitute $x = \sqrt{\pi}$ and $y = 1$ into the derivative:
$\frac{\text{d}y}{\text{d}x} = \frac{-2(\sqrt{\pi})(1)(1+1^2)}{(\sqrt{\pi})^2(1+1^2) - 1}$
$\frac{\text{d}y}{\text{d}x} = -\frac{4\sqrt{\pi}}{2\pi - 1}$