I would like to find the angle $\theta$ that the tangent to a curve $f(x)$ at a given point $(x,f(x))$ makes with the horizontal in terms of the coordinates $(x,f(x))$ of the point. See figure
Let $(v_{x},v_{y})$ be the tangent vector at $(x,f(x))$. Then we know that
$$\tan\theta=\frac{v_{y}}{v_{x}}$$
and also
$$\tan\phi=\frac{y}{x}$$
but how to write $\tan\theta$ in terms of $x$ and $y$?
The tangent velocity vector can be found by differentiating a paramatrisation $\langle x(t),y(t)\rangle$ of the curve (you can think of this like the position vector of a particle living on the curve).
Even if you don't like to find a good paramatrisation, we can just take the paramatrisation $x(t)=t$, in which case we have position vector $\langle x,y(x)\rangle$. The velocity vector is then $\langle 1,y'(x)\rangle$. Hence the angle between the velocity vector and the $x$ axis is the same as the angle between it and $\hat \imath$, the unit vector in the posiitve $x$ direction. The angle is calculated with the dot product as $$ \theta=\arccos\left(\frac{1}{\sqrt{1+(y'(x))^2}}\right)=\arctan(y'(x)). $$