Find the x-intercept and y-intercept of the curve. Determine the stationary points of the curve.

This gives two possible solutions:

$$x^2 = 0 \implies x = 0$$ $$3x + 4 = 0 \implies 3x = -4 \implies x = -\frac{4}{3}$$

The x-intercepts are $\boxed{x=0}$ and $\boxed{x=-\frac{4}{3}}$.

$$y = 0 + 0$$ $$y = 0$$

The y-intercept is $\boxed{y=0}$.

$$9x^2 + 8x = 0$$

Factor out $x$:

$$x(9x + 8) = 0$$

This gives two possible solutions for $x$:

$$x = 0$$ $$9x + 8 = 0 \implies 9x = -8 \implies x = -\frac{8}{9}$$

For $x = 0$:

$$y = 3(0)^3 + 4(0)^2 = 0$$

So, one stationary point is $(0,0)$.
For $x = -\frac{8}{9}$:

$$y = 3\left(-\frac{8}{9}\right)^3 + 4\left(-\frac{8}{9}\right)^2$$ $$y = 3\left(-\frac{512}{729}\right) + 4\left(\frac{64}{81}\right)$$ $$y = -\frac{1536}{729} + \frac{256}{81}$$

To add these fractions, find a common denominator (729):

$$y = -\frac{1536}{729} + \frac{256 \times 9}{81 \times 9}$$ $$y = -\frac{1536}{729} + \frac{2304}{729}$$ $$y = \frac{2304 - 1536}{729} = \frac{768}{729}$$

This fraction can be simplified by dividing the numerator and denominator by 3:

$$y = \frac{256}{243}$$

So, the other stationary point is $\left(-\frac{8}{9}, \frac{256}{243}\right)$.
The stationary points are $\boxed{(0,0)}$ and $\boxed{\left(-\frac{8}{9}, \frac{256}{243}\right)}$.

For the point $(0,0)$ (where $x=0$):

$$\frac{d^2y}{dx^2} = 18(0) + 8 = 8$$

Since $\frac{d^2y}{dx^2} > 0$, the point $(0,0)$ is a local minimum.
For the point $\left(-\frac{8}{9}, \frac{256}{243}\right)$ (where $x=-\frac{8}{9}$):

$$\frac{d^2y}{dx^2} = 18\left(-\frac{8}{9}\right) + 8$$ $$\frac{d^2y}{dx^2} = -2 \times 8 + 8$$ $$\frac{d^2y}{dx^2} = -16 + 8 = -8$$

Since $\frac{d^2y}{dx^2} < 0$, the point $\left(-\frac{8}{9}, \frac{256}{243}\right)$ is a local maximum.

The nature of the stationary points are:

• $(0,0)$ is a $\boxed{\text{local minimum}}$.
• $\left(-\frac{8}{9}, \frac{256}{243}\right)$ is a $\boxed{\text{local maximum}}$.