Spherical Coordinates 2

Fundementals of applied electromagnetics i am wondering where does that formula come from? (that i signed in pic.)

These are well-known properties of vectors, so if you are having trouble recognizing them, I really suggest a review. But here are the essentials:

At any point on a surface where the surface is smooth, there will be a plane tangent to the surface, and a line that normal to that plane (and therefore, to the surface).

So if you are given a vector $$\mathbf E$$ at that point, you can project $$\mathbf E$$ on to the normal line, to get the "normal component" $$\mathbf E_n$$, and you can also project it onto the tangent plane to get the "tangent component" $$\mathbf E_t$$. And because the normal line and tangent plane are orthogonal to each other,

$$\mathbf E = \mathbf E_t + \mathbf E_n$$

Now, if the direction of the normal line is given by the unit vector $$\mathbf{\hat n}$$, then the signed length of the projection of $$\mathbf E$$ is given by the inner product $$\bf E\cdot \hat n$$. And since lines are one dimensional, every vector on the line is a multiple of $$\mathbf{\hat n}$$. So it must be that $$\mathbf E_n = (\mathbf E\cdot \mathbf{\hat n})\mathbf{\hat n}$$

And therefore, the tangential component is given by

$$\mathbf E_t = \mathbf E - \mathbf E_n = \mathbf E - (\mathbf E\cdot \mathbf{\hat n})\mathbf{\hat n}$$

For a cylinder, the unit normal vector to the surface is the radial vector $$\mathbf{\hat r}$$, because the tangent to a circle is perpendicular to the radius.

For a sphere, the unit normal vector to the surface is the radial vector $$\hat {\boldsymbol\rho}$$, for exactly the same reason.