曲线 C 的方程为 $\dfrac{y^2}{x^2}=\dfrac{3a-x}{a+x},\quad x \ne 0,$ 其中 $a$ 是正常数。

1. Establish an inequality based on the non-negativity of a real square.

For real values of $x$ and $y$, the square term $\dfrac{y^2}{x^2}$ must be non-negative.
Therefore,
$\dfrac{3a-x}{a+x} \ge 0$

2. Solve the rational inequality algebraically.

Multiply both sides by $(a+x)^2$, noting that $x \ne -a$ to avoid division by zero:
$(3a-x)(a+x) \ge 0$
The critical values are $x = -a$ and $x = 3a$. Since the coefficient of $x^2$ is negative, the quadratic expression is non-negative between the roots:
$-a \le x \le 3a$
Since $x \ne -a$ (as it makes the original denominator zero), the inequality becomes:
$-a < x \le 3a$

3. Incorporate the given domain restriction to state the final range.

The question explicitly states $x \ne 0$.
Combining this restriction with the solved inequality, the range of values $x$ can take is:
$-a < x < 0 \quad \text{or} \quad 0 < x \le 3a$

1. Differentiate the curve equation implicitly with respect to $x$.

For $a=1$, the equation is $\dfrac{y^2}{x^2} = \dfrac{3-x}{1+x}$, which can be rewritten as:
$y^2(1+x) = 3x^2 - x^3$
Differentiating implicitly with respect to $x$:
$2y \frac{dy}{dx} (1+x) + y^2(1) = 6x - 3x^2$

2. Set $\frac{dy}{dx} = 0$ and solve for valid $x$-coordinates.

For the tangent to be parallel to the $x$-axis, $\frac{dy}{dx} = 0$.
Substituting this into the derivative equation gives:
$0 + y^2 = 6x - 3x^2$
Substitute $y^2 = 6x - 3x^2$ back into the rewritten curve equation $y^2(1+x) = 3x^2 - x^3$:
$(6x - 3x^2)(1+x) = 3x^2 - x^3$
$6x + 6x^2 - 3x^2 - 3x^3 = 3x^2 - x^3$
$6x + 3x^2 - 3x^3 = 3x^2 - x^3$
$6x - 2x^3 = 0$
$2x(3 - x^2) = 0$
This gives $x = 0$, $x = \sqrt{3}$, or $x = -\sqrt{3}$.

From part (a), the valid domain of the curve for $a=1$ is $-1 < x \le 3$ and $x \neq 0$. Therefore, we reject $x = 0$ and $x = -\sqrt{3}$, leaving $x = \sqrt{3}$ as the only valid solution.

3. Find the corresponding $y$-coordinates for $x = \sqrt{3}$.

Substitute $x = \sqrt{3}$ into our derived equation $y^2 = 6x - 3x^2$:
$y^2 = 6(\sqrt{3}) - 3(\sqrt{3})^2$
$y^2 = 6\sqrt{3} - 9$
$y = \pm\sqrt{6\sqrt{3} - 9}$
The required coordinates are $(\sqrt{3}, \sqrt{6\sqrt{3}-9})$ and $(\sqrt{3}, -\sqrt{6\sqrt{3}-9})$.

1. Find the $y$-coordinates on the curve when $x = 1$.

Substitute $x = 1$ and $a = 1$ into the curve equation:
$\frac{y^2}{1^2} = \frac{3(1) - 1}{1 + 1}$
$y^2 = \frac{2}{2} = 1$
$y = 1 \text{ or } y = -1$
So the points are $(1, 1)$ and $(1, -1)$.

2. Differentiate the equation implicitly with respect to $x$.

Using the quotient rule on both sides of $\frac{y^2}{x^2} = \frac{3-x}{1+x}$:
$\frac{2x^2 y \frac{dy}{dx} - 2xy^2}{x^4} = \frac{(-1)(1+x) - (3-x)(1)}{(1+x)^2}$
$\frac{2xy \frac{dy}{dx} - 2y^2}{x^3} = \frac{-4}{(1+x)^2}$

3. Evaluate the gradients of the normals at $(1, 1)$ and $(1, -1)$.

At $(1, 1)$:
$\frac{2(1)(1) \frac{dy}{dx} - 2(1)^2}{1^3} = \frac{-4}{(1+1)^2}$
$2 \frac{dy}{dx} - 2 = -1 \implies \frac{dy}{dx} = \frac{1}{2}$
The gradient of the normal is $m_1 = -\frac{1}{1/2} = -2$.

At $(1, -1)$:
$\frac{2(1)(-1) \frac{dy}{dx} - 2(-1)^2}{1^3} = \frac{-4}{(1+1)^2}$
$-2 \frac{dy}{dx} - 2 = -1 \implies \frac{dy}{dx} = -\frac{1}{2}$
The gradient of the normal is $m_2 = -\frac{1}{-1/2} = 2$.

4. Form the equations of the normals and find their point of intersection.

Equation of normal at $(1, 1)$:
$y - 1 = -2(x - 1) \implies y = -2x + 3$

Equation of normal at $(1, -1)$:
$y - (-1) = 2(x - 1) \implies y = 2x - 3$

Equating the two normals to find intersection $N$:
$-2x + 3 = 2x - 3$
$4x = 6 \implies x = \frac{3}{2}$
Substitute $x = \frac{3}{2}$ into $y = 2x - 3$:
$y = 2\left(\frac{3}{2}\right) - 3 = 0$

Therefore, the coordinates of $N$ are $\left(\frac{3}{2}, 0\right)$.