I have homework on slope fields where I have to graph a bunch and find the equillibrium solution, but instead of taking such a long time to graph them, I decided to use WolframAlpha. Sadly, there is no function to generate a slope field for you, and the closest alternative I found is a vector field.
Now my understanding of a vector field is that it graphs the vector $\{f(x, y), g(x, y)\}$ for all $(x, y)$ points in the coordinate plane. From here I found that given a slope field $y' = f(x, y)$, you can graph the slope field as a vector field where each vector has a magnitude of $1$ and slope $f(x, y)$ for all $(x, y)$ points.
Now as an example, I tried to graph the slope field $y' = x + y$ as a vector field. I let $$y' = x + y = m$$ where $m$ is the value you get by plugging a point into the differential equation. Then I said we have to graph the vector $\{a, b\}$ at point $(x, y)$ such that the slope of the vector is $x+y$ and its magnitude is $1$. So $$a^2+b^2 = 1$$ $$b/a = x+y = m$$ And $$a = \frac{1}{\sqrt{m^2+1}}, b = \frac{m}{\sqrt{m^2+1}}$$
Now my question is, can this be extended to all slope fielda, even ones involving trigonometric expressions and logs and exponents?
Note that $f$ in the vector $\langle f(x,y), g(x,y)\rangle$ and the $f(x,y)$ in the linked article are not the same $f$. Your notation gives the vector field in terms of its components, which is a perfectly good way to describe it. The linked article uses $f(x,y)$ to express the slope of the vector field at a point $(x,y)$, which is also fine as long as the vector field is not vertical (in which case your $f(x,y)$ is zero, and the slope is infinite).
If you are given a vector field $\langle f(x,y), g(x,y)\rangle$, then the slope field is the same field except that you say you want unit vectors. So the answer is simply $$\frac{1}{\sqrt{f(x,y)^2+g(x,y)^2}}\langle f(x,y), g(x,y)\rangle$$ unless $f(x,y) = g(x,y) = 0$, in which case the answer is $\langle 0, 0\rangle$.
If you are given a vector field in the form $y' = f(x,y)$, you can write this as the vector field $\langle 1, f(x,y)\rangle$ (that is, a unit change in $x$ produces a change of $y'$ in $y$). Then by the method above, you get for the unit vector in this direction $$\frac{1}{\sqrt{f(x,y)^2+1}}\langle 1,f(x,y)\rangle$$ as you said. Note that when you are given a vector field in this form you do not have to worry about the slope being vertical since $y'$ is not defined in this case.