I'm new to this community, so please bear with me if I make a mistake following rules and conventions, and thanks in advance.
I'm currently learning - following along from this video ("Implicit differentiation, what's going on here? | Essence of calculus").
At 2:42 in the video timeline, the teacher mentions that the implicit curve equation $x^2+y^2=5^2$, which is then resolved into its derivative form: $2x\,\mathrm{d}x+2y\,\mathrm{d}y=0$ can then be rearranged to the form expression: $\frac{\mathrm dy}{\mathrm dx} = -\frac{x}{y}$. I wanted to understand how this rearrangement happened? I tried to explore this equation rearrangement: $2x\,\mathrm{d}x+2y\,\mathrm{d}y=0$ to $\frac{\mathrm dy}{\mathrm dx} = -\frac{x}{y}$ from online sources, but to no joy. By the way, I also tried to make the question more readable in terms of using the proper maths equation symbols syntax, using this link, but I got an error message from this website saying that I don't have enough reputation score to post images in my question. So I hope I made this question as readable and clear as possible, with the limits I'm given. Thanks in advance.
As noted in the comments, the method of implicit differentiation that Sanderson presents is an abuse of notation used provide a more intuitive understanding of why it works. A more formal process would resemble the following \begin{align} x^2+y^2=5&\implies \frac{\mathrm{d}}{\mathrm{d}x}(x^2+y^2)=\frac{\mathrm{d}}{\mathrm{d}x}5^2 \\ &\implies 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0 \implies 2y\frac{\mathrm{d}y}{\mathrm{d}x}=−2x \\ &\implies \frac{\mathrm{d}y}{\mathrm{d}x}=−\frac{x}{y} \end{align}