I'll have to figure out the tangent of a curve given by the expression $f(x,y)=constant$ and the angle $\theta$ it makes at a given point $(x=a,y=b)$. What I am getting is that the tan of the angle is being reproduced by
$$\tan(\theta)=-\frac{\partial_x f}{\partial_y f}\Bigg |_{(a,b)}$$
however, I don't really understand why that must be true. Can someone kindly prove this result or explain how to go about getting a general expression for the tangent of the given curve at a given point?
Let $L:=\{(x,y)\in \mathbb{R}^2:f(x,y)=K\}$. If $g:(-1,1)\to L$ is a (local) regular parametrization of $L$ with $g(0)=(a,b)$ then a tangent vector at $(a,b)$ in $\operatorname{img}(g)$ is given by $g'(0)$. Now observe that $f\circ g=K$, therefore
$$ \partial (f\circ g)(0)=\partial f(g(0))g'(0)=\nabla f(a,b)\cdot g'(0)=0 $$
so any tangent vector at $(a,b)$ in $\operatorname{img}(g)$ is orthogonal to $\nabla f(a,b)$. From here you can easily find a normalized tangent vector at $(a,b)$ just rotating $v:=\frac{\nabla f(a,b)}{\|\nabla f(a,b)\|}$ 90 degrees clockwise, from where compute your angle $\theta $.