曲线 $C$ 的方程为 $x^{3} + y^{3} - 5xy = 0$，其中 $x>0$。

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

$\frac{d}{dx}(x^3 + y^3 - 5xy) = 0$
$3x^2 + 3y^2 \frac{dy}{dx} - \left( 5y + 5x \frac{dy}{dx} \right) = 0$

2. Group the $\frac{dy}{dx}$ terms and make it the subject.

$3y^2 \frac{dy}{dx} - 5x \frac{dy}{dx} = 5y - 3x^2$
$\frac{dy}{dx}(3y^2 - 5x) = 5y - 3x^2$
$\frac{dy}{dx} = \frac{5y - 3x^2}{3y^2 - 5x}$

1. Establish the relationship between $x$ and $y$ using the condition for the normal.

From part (a), we have the derivative $\frac{dy}{dx} = \frac{5y - 3x^2}{3y^2 - 5x}$.

If the normal is parallel to the $y$-axis (a vertical line), the tangent to the curve at that point must be perpendicular to it, meaning the tangent is parallel to the $x$-axis (a horizontal line).
Therefore, the gradient of the tangent is zero:
$\frac{dy}{dx} = 0$
$\frac{5y - 3x^2}{3y^2 - 5x} = 0$
$5y - 3x^2 = 0$
$y = \frac{3}{5}x^2$

2. Substitute the relationship into the curve's equation to solve for $x$.

Substitute $y = \frac{3}{5}x^2$ into the equation of the curve $x^3 + y^3 - 5xy = 0$:
$x^3 + \left(\frac{3}{5}x^2\right)^3 - 5x\left(\frac{3}{5}x^2\right) = 0$
$x^3 + \frac{27}{125}x^6 - 3x^3 = 0$
$\frac{27}{125}x^6 - 2x^3 = 0$
$x^3 \left( \frac{27}{125}x^3 - 2 \right) = 0$
Since it is given that $x > 0$, we know $x \neq 0$, allowing us to divide by $x^3$:
$\frac{27}{125}x^3 = 2$
$x^3 = \frac{250}{27}$
$x = \sqrt[3]{\frac{250}{27}} = \frac{\sqrt[3]{125 \times 2}}{3} = \frac{5\sqrt[3]{2}}{3}$

3. Calculate the corresponding $y$-coordinate and state the final point.

Substitute $x = \frac{5\sqrt[3]{2}}{3}$ back into $y = \frac{3}{5}x^2$:
$y = \frac{3}{5} \left( \frac{5\sqrt[3]{2}}{3} \right)^2$
$y = \frac{3}{5} \left( \frac{25 \times 2^{2/3}}{9} \right)$
$y = \frac{5 \times 2^{2/3}}{3} = \frac{5\sqrt[3]{4}}{3}$

The coordinates of the point are $\left( \frac{5\sqrt[3]{2}}{3}, \frac{5\sqrt[3]{4}}{3} \right)$.

1. Concisely retrieve the coordinates of the stationary point (required from part (b) to complete this target).

At the stationary point (where the normal is parallel to the $y$-axis), the tangent is horizontal, so $\frac{dy}{dx} = 0$.
From part (a), $\frac{dy}{dx} = \frac{5y - 3x^2}{3y^2 - 5x}$. Setting this to $0$ gives $5y = 3x^2 \implies y = \frac{3}{5}x^2$.
Substituting $y = \frac{3}{5}x^2$ into $x^3 + y^3 - 5xy = 0$:
$x^3 + \left(\frac{3}{5}x^2\right)^3 - 5x\left(\frac{3}{5}x^2\right) = 0 \implies x^3 + \frac{27}{125}x^6 - 3x^3 = 0$
$x^3\left(\frac{27}{125}x^3 - 2\right) = 0$
Since $x > 0$, $x^3 = \frac{250}{27} \implies x = \frac{5\sqrt[3]{2}}{3}$.
Consequently, $y = \frac{3}{5}\left(\frac{5\sqrt[3]{2}}{3}\right)^2 = \frac{5\sqrt[3]{4}}{3}$.

2. Find an expression for the second derivative $\frac{d^2y}{dx^2}$ at the stationary point.

Using the quotient rule on $\frac{dy}{dx} = \frac{5y - 3x^2}{3y^2 - 5x}$:
$\frac{d^2y}{dx^2} = \frac{(3y^2 - 5x)\left(5\frac{dy}{dx} - 6x\right) - (5y - 3x^2)\left(6y\frac{dy}{dx} - 5\right)}{(3y^2 - 5x)^2}$
At the stationary point, we know $\frac{dy}{dx} = 0$ and the numerator of the first derivative is $5y - 3x^2 = 0$. Substituting these directly:
$\frac{d^2y}{dx^2} = \frac{(3y^2 - 5x)(-6x) - 0}{(3y^2 - 5x)^2} = -\frac{6x}{3y^2 - 5x}$

3. Evaluate the second derivative to determine the nature of the stationary point.

Substitute $x = \frac{5\sqrt[3]{2}}{3}$ and $y = \frac{5\sqrt[3]{4}}{3}$ into the second derivative expression:
$3y^2 - 5x = 3\left(\frac{5\sqrt[3]{4}}{3}\right)^2 - 5\left(\frac{5\sqrt[3]{2}}{3}\right) = 3\left(\frac{25 \cdot 2\sqrt[3]{2}}{9}\right) - \frac{25\sqrt[3]{2}}{3} = \frac{50\sqrt[3]{2}}{3} - \frac{25\sqrt[3]{2}}{3} = \frac{25\sqrt[3]{2}}{3}$
$\frac{d^2y}{dx^2} = \frac{-6\left(\frac{5\sqrt[3]{2}}{3}\right)}{\frac{25\sqrt[3]{2}}{3}} = \frac{-10\sqrt[3]{2}}{\frac{25\sqrt[3]{2}}{3}} = -\frac{30}{25} = -1.2$
Since $\frac{d^2y}{dx^2} < 0$, the stationary point is a maximum.