Find the first derivative of the function.

The given function is $y = \frac{1}{3}x^3 - x + 7$.
To find the turning points, we first calculate the first derivative $\frac{dy}{dx}$.

$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{3}x^3 - x + 7\right)$$ $$\frac{dy}{dx} = \frac{1}{3}(3x^2) - 1 + 0$$ $$\frac{dy}{dx} = x^2 - 1$$

$$x^2 - 1 = 0$$ $$x^2 = 1$$ $$x = \pm\sqrt{1}$$

So, $x = 1$ or $x = -1$. These are the x-coordinates of the turning points.

Substitute the x-values back into the original function $y = \frac{1}{3}x^3 - x + 7$.

For $x = 1$:

$$y = \frac{1}{3}(1)^3 - (1) + 7$$ $$y = \frac{1}{3} - 1 + 7$$ $$y = \frac{1}{3} + 6$$ $$y = \frac{1}{3} + \frac{18}{3}$$ $$y = \frac{19}{3}$$

The first turning point is $\left(1, \frac{19}{3}\right)$.

For $x = -1$:

$$y = \frac{1}{3}(-1)^3 - (-1) + 7$$ $$y = \frac{1}{3}(-1) + 1 + 7$$ $$y = -\frac{1}{3} + 8$$ $$y = -\frac{1}{3} + \frac{24}{3}$$ $$y = \frac{23}{3}$$

The second turning point is $\left(-1, \frac{23}{3}\right)$.

The first derivative is $\frac{dy}{dx} = x^2 - 1$.

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(x^2 - 1)$$ $$\frac{d^2y}{dx^2} = 2x$$

For $x = 1$:

$$\frac{d^2y}{dx^2} = 2(1) = 2$$

Since $\frac{d^2y}{dx^2} > 0$, the turning point $\left(1, \frac{19}{3}\right)$ is a local minimum.

For $x = -1$:

$$\frac{d^2y}{dx^2} = 2(-1) = -2$$

Since $\frac{d^2y}{dx^2} < 0$, the turning point $\left(-1, \frac{23}{3}\right)$ is a local maximum.

The turning points are:
Local minimum: $\boxed{\left(1, \frac{19}{3}\right)}$
Local maximum: $\boxed{\left(-1, \frac{23}{3}\right)}$