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 $x$, applying the product rule to the $xy$ term.

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

2. Isolate $\dfrac{dy}{dx}$ by collecting relevant terms on one side of the equation.

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

1. Find the expression for $\frac{dy}{dx}$ by differentiating the equation of the curve implicitly (as required from part a).

$x^3 + y^3 - 5xy = 0$
Differentiating implicitly with respect to $x$:
$3x^2 + 3y^2 \frac{dy}{dx} - 5 \left( y + x\frac{dy}{dx} \right) = 0$
$3x^2 + 3y^2 \frac{dy}{dx} - 5y - 5x\frac{dy}{dx} = 0$
$(3y^2 - 5x)\frac{dy}{dx} = 5y - 3x^2$
$\frac{dy}{dx} = \frac{5y - 3x^2}{3y^2 - 5x}$

2. Relate the given normal condition to the gradient of the tangent.

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

3. Substitute the relationship between $x$ and $y$ into the original equation to solve for $x$.

Substitute $y = \frac{3}{5}x^2$ into $x^3 + y^3 - 5xy = 0$:
$x^3 + \left(\frac{3}{5}x^2\right)^3 - 5x\left(\frac{3}{5}x^2\right) = 0$
$x^3 + \frac{27}{125}x^6 - 3x^3 = 0$
$\frac{27}{125}x^6 = 2x^3$
Since $x > 0$, we can divide both sides by $x^3$:
$\frac{27}{125}x^3 = 2$
$x^3 = \frac{250}{27}$
$x = \frac{\sqrt[3]{250}}{3} = \frac{5\sqrt[3]{2}}{3} \approx 2.10$

4. Find the corresponding $y$-coordinate to state the final point.

Substitute $x$ back to find $y$:
$y = \frac{3}{5}\left(\frac{5\sqrt[3]{2}}{3}\right)^2 = \frac{3}{5} \left(\frac{25\sqrt[3]{4}}{9}\right) = \frac{5\sqrt[3]{4}}{3} \approx 2.65$
The coordinates are $\left( \frac{5\sqrt[3]{2}}{3}, \frac{5\sqrt[3]{4}}{3} \right)$.

1. Find an expression for the second derivative $\frac{d^2y}{dx^2}$ at the stationary point by differentiating the result from part (a).

From part (a), the first derivative is:
$\frac{dy}{dx} = \frac{5y - 3x^2}{3y^2 - 5x}$
Using the quotient rule to find the second derivative:
$\frac{d^2y}{dx^2} = \frac{(3y^2 - 5x) \frac{d}{dx}(5y - 3x^2) - (5y - 3x^2) \frac{d}{dx}(3y^2 - 5x)}{(3y^2 - 5x)^2}$
$\frac{d^2y}{dx^2} = \frac{(3y^2 - 5x)\left(5\frac{dy}{dx} - 6x\right) - (5y - 3x^2)\left(6y\frac{dy}{dx} - 5\right)}{(3y^2 - 5x)^2}$
In part (b), we found the point where the normal is parallel to the $y$-axis. This implies the tangent is parallel to the $x$-axis, meaning it is a stationary point where $\frac{dy}{dx} = 0$ and $5y - 3x^2 = 0$.
Substituting these into the second derivative expression heavily simplifies it:
$\frac{d^2y}{dx^2} = \frac{(3y^2 - 5x)(0 - 6x) - (0)}{(3y^2 - 5x)^2} = \frac{-6x}{3y^2 - 5x}$

2. Evaluate the second derivative using the coordinates of the stationary point to determine its nature.

From part (b), the coordinates of the stationary point are $x = \frac{5\sqrt[3]{2}}{3}$ and $y = \frac{5\sqrt[3]{4}}{3}$.
Evaluate the denominator, $3y^2 - 5x$:
$3y^2 - 5x = 3\left(\frac{5\sqrt[3]{4}}{3}\right)^2 - 5\left(\frac{5\sqrt[3]{2}}{3}\right)$
$= 3\left(\frac{25 \times 2\sqrt[3]{2}}{9}\right) - \frac{25\sqrt[3]{2}}{3} = \frac{50\sqrt[3]{2}}{3} - \frac{25\sqrt[3]{2}}{3} = \frac{25\sqrt[3]{2}}{3}$
Now, evaluate the second derivative:
$\frac{d^2y}{dx^2} = \frac{-6\left(\frac{5\sqrt[3]{2}}{3}\right)}{\frac{25\sqrt[3]{2}}{3}} = \frac{-10\sqrt[3]{2}}{\frac{25\sqrt[3]{2}}{3}} = -10 \times \frac{3}{25} = -\frac{6}{5}$
Since $\frac{d^2y}{dx^2} = -1.2 < 0$, the stationary point is a local maximum.