方程为 $y = \mathrm{f}(x)$ 的曲线在点 $(x, y)$ 处的切线通过点 $(a, x)$，其中 $a$ 是正常数且 $x < a$。

1. Express the gradient of the tangent using the given points

The gradient of the tangent to the curve at the point $(x, y)$ is given by the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$.

Since the tangent line also passes through the point $(a, x)$, we can calculate its gradient using the two-point formula $m = \frac{y_2 - y_1}{x_2 - x_1}$:
$\text{Gradient} = \frac{x - y}{a - x}$

2. Equate the two expressions to form the differential equation

Equating the two expressions for the gradient gives the differential equation:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x - y}{a - x}$

1. Determine the differential equation from part (i) to establish the starting point.

The tangent at $(x,y)$ passes through $(a,x)$. The gradient of this tangent is $\frac{\mathrm{d}y}{\mathrm{d}x}$.
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x - y}{a - x}$

2. Differentiate the given substitution $y = (a - x)z$ with respect to $x$.

Using the product rule:
$\frac{\mathrm{d}y}{\mathrm{d}x} = -z + (a - x)\frac{\mathrm{d}z}{\mathrm{d}x}$

3. Substitute into the differential equation to show the transformed equation.

Substitute $y = (a - x)z$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$ into the differential equation:
$-z + (a - x)\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{x - (a - x)z}{a - x}$
$-z + (a - x)\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{x}{a - x} - z$
$(a - x)\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{x}{a - x}$
$\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{x}{(a - x)^2} \quad \text{(shown)}$

4. Express the right hand side using partial fractions.

Let $\frac{x}{(a - x)^2} = \frac{A}{a - x} + \frac{B}{(a - x)^2}$
$x = A(a - x) + B$
Substituting $x = a$ gives $B = a$.
Comparing coefficients of $x$ gives $-A = 1 \implies A = -1$.
Thus,
$\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{-1}{a - x} + \frac{a}{(a - x)^2}$

5. Integrate to find $z$.

$z = \int \left( \frac{-1}{a - x} + \frac{a}{(a - x)^2} \right) \,\mathrm{d}x$
$z = \ln(a - x) + \frac{a}{a - x} + C$
(Note: Since $x < a$, $a - x > 0$, so no absolute value is required for the natural logarithm.)

6. Substitute $z$ back to find the general solution for $y$.

Since $z = \frac{y}{a - x}$,
$\frac{y}{a - x} = \ln(a - x) + \frac{a}{a - x} + C$
$y = (a - x)\ln(a - x) + a + C(a - x)$
where $C$ is an arbitrary constant.

1. First, derive the general solution for $y$ from the differential equation in part (ii).

$\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{x}{(a - x)^2} = \frac{-(a - x) + a}{(a - x)^2} = -\frac{1}{a - x} + \frac{a}{(a - x)^2}$
$z = \int \left( -\frac{1}{a - x} + \frac{a}{(a - x)^2} \right) \mathrm{d}x$
$z = \ln(a - x) + \frac{a}{a - x} + c$
$y = (a - x)z = (a - x)\ln(a - x) + a + c(a - x)$

2. Use the given initial condition (the origin) to find the constant of integration $c$.

Given that the curve passes through $(0, 0)$:
$0 = (a - 0)\ln(a - 0) + a + c(a - 0)$
$0 = a\ln(a) + a + ca$
Since $a > 0$, we can divide the equation by $a$:
$0 = \ln(a) + 1 + c$
$c = -1 - \ln(a)$

3. Substitute $c$ back into the general solution and simplify using logarithm properties to obtain the required expression.

$\mathrm{f}(x) = (a - x)\ln(a - x) + a + (-1 - \ln(a))(a - x)$
$\mathrm{f}(x) = (a - x)\ln(a - x) + a - (a - x) - (a - x)\ln(a)$
$\mathrm{f}(x) = (a - x)[\ln(a - x) - \ln(a)] + a - a + x$
$\mathrm{f}(x) = (a - x)\ln\left(\frac{a - x}{a}\right) + x$