已知一条曲线的方程为 $y = \frac{2x - 3}{3x + 4}$，其中 $x \ne -\frac{4}{3}$。
曲线在点 $P$（其中 $x > 0$）处的法线平行于直线 $y = -\frac{25}{17}x + 2$。求点 $P$ 的坐标。 证明函数 $y = \frac{4}{3}x^3 - x^2 + 3x - 10$ 对于所有实数 $x$ 始终是单调递增的。

1. Find the gradient of the tangent from the given normal line.

The normal is parallel to $y = -\frac{25}{17}x + 2$, so the gradient of the normal is $-\frac{25}{17}$.
Let the gradient of the tangent be $m$.
$m \times \left(-\frac{25}{17}\right) = -1$
$m = \frac{17}{25}$

2. Differentiate the equation of the curve to find the gradient function.

$y = \frac{2x - 3}{3x + 4}$
Using the quotient rule:
$\frac{dy}{dx} = \frac{2(3x + 4) - 3(2x - 3)}{(3x + 4)^2}$
$\frac{dy}{dx} = \frac{6x + 8 - 6x + 9}{(3x + 4)^2}$
$\frac{dy}{dx} = \frac{17}{(3x + 4)^2}$

3. Equate the gradient function to the gradient of the tangent and solve for $x$.

$\frac{17}{(3x + 4)^2} = \frac{17}{25}$
$(3x + 4)^2 = 25$
$3x + 4 = 5 \quad \text{or} \quad 3x + 4 = -5$
$3x = 1 \quad \text{or} \quad 3x = -9$
$x = \frac{1}{3} \quad \text{or} \quad x = -3$
Since $x > 0$, we reject $x = -3$ and take $x = \frac{1}{3}$.

4. Find the corresponding $y$-coordinate of $P$.

Substitute $x = \frac{1}{3}$ into the equation of the curve:
$y = \frac{2(\frac{1}{3}) - 3}{3(\frac{1}{3}) + 4}$
$y = \frac{\frac{2}{3} - 3}{1 + 4}$
$y = \frac{-\frac{7}{3}}{5} = -\frac{7}{15}$
The coordinates of $P$ are $\left(\frac{1}{3}, -\frac{7}{15}\right)$.

1. Find the first derivative of the function.

$\frac{dy}{dx} = \frac{d}{dx}(\frac{4}{3}x^3 - x^2 + 3x - 10)$
$\frac{dy}{dx} = 4x^2 - 2x + 3$

2. Complete the square for the derivative to show it is always positive.

$\frac{dy}{dx} = 4(x^2 - \frac{1}{2}x) + 3$
$= 4[(x - \frac{1}{4})^2 - (\frac{1}{4})^2] + 3$
$= 4(x - \frac{1}{4})^2 - 4(\frac{1}{16}) + 3$
$= 4(x - \frac{1}{4})^2 - \frac{1}{4} + 3$
$= 4(x - \frac{1}{4})^2 + \frac{11}{4}$

3. Conclude that the derivative is always positive, hence the function is always increasing.

Since $(x - \frac{1}{4})^2 \ge 0$ for all real $x$,
$4(x - \frac{1}{4})^2 + \frac{11}{4} \ge \frac{11}{4} > 0$.
Since $\frac{dy}{dx} > 0$ for all real $x$, the function $y$ is always increasing.