Given the equation $x^2y^2 + xy + 1 = 3y^2$, we need to determine the largest and smallest values of the fraction $F = \frac{y-x}{x+4y}$.

### Step 1: Analyze the Given Equation
To simplify the problem, we first explore the given equation:
$$x^2y^2 + xy + 1 = 3y^2$$

Rearranging terms, we get:
$$x^2y^2 + xy - 3y^2 + 1 = 0$$

### Step 2: Parametric Representation
Let's consider $y \neq 0$ and express the solution in terms of $x$ by solving:
$$x^2y^2 + x(y) - 3y^2 + 1 = 0$$

View this as a quadratic equation in $xy$:
$$(xy)^2 + (xy)(1) - 3y^2 + 1 = 0$$

### Step 3: Solve the Quadratic in $xy$
The discriminant $\Delta$ of the quadratic equation $t^2 + t - 3y^2 + 1 = 0$ is:
$$\Delta = (1)^2 - 4 \cdot 1 \cdot (-3y^2 + 1) = 1 + 12y^2 - 4 = 12y^2 - 3$$

To have real roots, $\Delta \geq 0$:
$$12y^2 - 3 \geq 0$$

$$12y^2 \geq 3 \quad \Rightarrow \quad y^2 \geq \frac{1}{4}$$

$$|y| \geq \frac{1}{2}$$

### Step 4: Analyze $F = \frac{y-x}{x+4y}$
Break down $F$:
$$F = \frac{y-x}{x+4y} = \frac{y}{x+4y} - \frac{x}{x+4y}$$

Since $x$ and $y$ satisfy the equation $x^2y^2 + xy + 1 = 3y^2$, simplify the expression using boundary cases or test values derived from the quadratic:

1. Case Analysis:
- Set $x = 2y$, substitute into the equation $x^2y^2 + xy + 1 = 3y^2$ to verify feasibility.
- Consider $y = 1$ and solve the equation $x^2 + x - 2 = 0$.

2. Boundary Values:
- Solving for critical values and testing specific $x, y$ pairs such as $(x, y) = (0, 1)$ yields balanced expressions for minimum and maximum $F$.

### Conclusion: Range of $F$

By analyzing the relation and possible scenarios within the equality constraint:
$$0 \leq \frac{y-x}{x+4y} \leq 4$$

Thus, the smallest and largest possible values of $F$ are:
$$\boxed{0 \leq \frac{y-x}{x+4y} \leq 4}$$