To solve, we convert the equation of the parabola into its standard form, obtaining $x^{2}=-\frac{1}{4}y$.
From this, we find that $2p=\frac{1}{4}$, hence $p=\frac{1}{8}$, and $\frac{p}{2}=\frac{1}{16}$.
Knowing that the parabola opens downwards, the equation of the directrix is: $y=\frac{p}{2}=\frac{1}{16}$.
Therefore, the answer is: $y=\frac{1}{16}$
By converting the equation of the parabola into its standard form, we can find the value of $p$, and combining this with the direction in which the parabola opens, we can determine the equation.
This question tests the basic properties of parabolas, involving the solution for the equation of the directrix, and is considered a basic question.

Thus, the equation of the directrix of the parabola is $\boxed{y=\frac{1}{16}}$.

Rearrange the equation of the parabola to get \x^{2}= \dfrac {1}{4}y\, thus \p= \dfrac {1}{8}\.
Since the parabola opens upwards,
the equation of the directrix is \y=- \dfrac {1}{16}\.
Therefore, the answer is: \y=- \dfrac {1}{16}\.
First, rearrange the equation of the parabola into the standard form, then find \p\, and finally, use the properties of the parabola to determine the equation of the directrix.
This question mainly examines the standard equation of a parabola and its simple properties. It is a basic question.

$\boxed{y=- \dfrac {1}{16}}$

Analysis

This question examines the standard equation of a parabola and its geometric properties. Identifying the type and positioning is key. First, we convert the parabola \y=4x^2\ into its standard form, \x^2= \frac{1}{4}y\, which indicates that the focus is on the positive half of the y-axis and the parabola opens upwards. From \2p= \frac{1}{4}\, we can determine the equation of the directrix of the parabola \y=4x^2\.

Solution

Given that the parabola \y=4x^2\ can be rewritten as \x^2= \frac{1}{4}y\, with the focus on the positive half of the y-axis and opening upwards, and \2p= \frac{1}{4}\,

It follows that \\frac{p}{2}= \frac{1}{16}\,

Therefore, the equation of the directrix of the parabola \y=4x^2\ is \y= -\frac{1}{16}\,

Hence, the answer is $\boxed{y= -\frac{1}{16}}$.