Definitions
Unit vector has length 1. Orthonormal vectors are orthogonal and unit vectors.
RobJohn's suggestions for the basis in polar coordinates, here, satisfy the criteria but how can you approach this geometrically? And how to make sure they are actually orthogonal in polar coordinates?
$$\begin{align} \hat{e}_R&=\frac{(x,y,z)}{r}\\ \hat{e}_\theta&=\frac{(-y,x,0)}{\sqrt{r^2-z^2}}\\ \hat{e}_\phi&=\frac{(-xz,-yz,r^2-z^2)}{r\sqrt{r^2-z^2}} \end{align}$$
Perhaps related
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Visual Ways to Remember Cross products of Unit vectors? Cross-product in $\mathbb F^3$? (third vector from two other unit vectors)
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Explain Dot product with Partial derivatives in Polar-coordinates (orthogonality check)
Example problem (about page 817 here)
$$\nabla\cdot\bar{F}=\left(\hat{e}_{R}\partial_{R}+\frac{1}{R}\hat{e}_{\theta}\partial_{\theta}+
\frac{1}{R\sin(\theta)}
\hat{e}_{\phi}\partial_{\phi}\right)\cdot\bar{F}$$where
$$\bar{F}=R^3 (
\cos(\phi)\sin(\theta)\bar{i}+\sin(\phi)\sin(\theta)\bar{j}+\cos(\theta)\bar{k}).$$
Best Answer
$\hat{e}_R$ is the unit radial vector. This is simply $(x,y,z)$ divided by its length, $r$: $\hat{e}_R=\frac{(x,y,z)}{r}$.
$\hat{e}_\theta$ is the unit vector tangent to the sphere, thus perpendicular to $(x,y,z)$, which is also perpendicular to $(0,0,1)$ since changing $\theta$ does not change $z$. Therefore, we simply need to take the cross product of $(x,y,z)$ and $(0,0,1)$ and normalize to get $\hat{e}_\theta=\frac{(-y,x,0)}{\sqrt{r^2-z^2}}$. The sign is chosen so that the vector points counter-clockwise.
$\hat{e}_\phi$ is perpendicular to the other two, so we take the cross product and normalize: $\hat{e}_\phi=\frac{(-xz,-yz,r^2-z^2)}{r\sqrt{r^2-z^2}}$. Again, the sign is chosen to have positive $z$ component.
Thus, we get $$ \begin{align} \hat{e}_R&=\frac{(x,y,z)}{r}\\ \hat{e}_\theta&=\frac{(-y,x,0)}{\sqrt{r^2-z^2}}\\ \hat{e}_\phi&=\frac{(-xz,-yz,r^2-z^2)}{r\sqrt{r^2-z^2}} \end{align} $$
References