A curve $C$ has equation $x^2 y + y^4 = 5$.

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

$\frac{\mathrm{d}}{\mathrm{d}x}(x^2 y + y^4) = \frac{\mathrm{d}}{\mathrm{d}x}(5)$
Using the product rule for the $x^2y$ term and the chain rule for the $y^4$ term:
$x^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy + 4y^3 \frac{\mathrm{d}y}{\mathrm{d}x} = 0$

2. Rearrange the terms to make $\frac{\mathrm{d}y}{\mathrm{d}x}$ the subject.

Factor out $\frac{\mathrm{d}y}{\mathrm{d}x}$ on the left side:
$(x^2 + 4y^3)\frac{\mathrm{d}y}{\mathrm{d}x} = -2xy$
Divide both sides by $(x^2 + 4y^3)$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2xy}{x^2 + 4y^3}$

1. Use the implicit derivative from part (a) to establish the condition for horizontal tangents.

From part (a), we have $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2xy}{x^2 + 4y^3}$.
For tangents parallel to the $x$-axis, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$.
$\frac{-2xy}{x^2 + 4y^3} = 0$
$-2xy = 0 \implies x = 0 \text{ or } y = 0$

2. Substitute the possible values into the curve's equation to find the corresponding $y$-coordinates.

If $y = 0$, substituting into the curve's equation $x^2y + y^4 = 5$ gives $0 = 5$, which is impossible. Thus, $y \neq 0$.
Substituting $x = 0$ into the curve's equation:
$0^2(y) + y^4 = 5$
$y^4 = 5$
$y = \pm\sqrt[4]{5}$

3. Calculate the exact distance between the two horizontal tangents.

The tangents are horizontal lines given by the equations $y = \sqrt[4]{5}$ and $y = -\sqrt[4]{5}$.
Exact distance $= \sqrt[4]{5} - (-\sqrt[4]{5})$
$= 2\sqrt[4]{5}$