Lets say we have $x^2 + y^2 = 1$
If we apply the derivative operator with respect to $x$ to both sides of the equation and solve for it we get: $\frac{dy}{dx}=\frac{-x}{y}$
I asked myself, what if we applied the derivative operator to equation with respect to y: well we get $\frac{dx}{dy}= \frac{-y}{x}$
All i want to know is if this is still viable to find the the slope of the tangent line at any given point within the equation.
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Yes, you can.
Let's do an example of this. The point $(\frac{\sqrt{3}}{2},\frac{1}{2})$ belongs to this curve, so, we can find the slope of the tangent line to the curve in this point by using the formula you previously stated ($y'=\frac{-x}{y}$).
Solving for this point: $$y'=-\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\\y'=-\sqrt{3}$$ This would be the desired slope.
But what happens when you try to get this result via usual differentiation? Near the given point, the curve can be written as the function $f(x)=\sqrt{1-x^2}$. If we derive normally respect to x, we obtain $$f'(x)=-\frac{x}{\sqrt{1-x^2}}$$ And solving for the given point, $$f'(\frac{\sqrt{3}}{2})=-\frac{\frac{\sqrt{3}}{2}}{\sqrt{1-\frac{3}{4}}}\\=-\frac{\sqrt{3}}{2\sqrt{\frac{1}{4}}}\\=-\frac{\sqrt{3}}{2(\frac{1}{2})}\\=-\sqrt{3}\\\therefore\\f'(\frac{\sqrt{3}}{2})=-\sqrt{3}$$ This demonstrates the previously stated method. Notice that on the $f'(x)$ formula, if you let $y=\sqrt{1-x²}$, you get your original formula for the slope.
Yes, of course, you can find the slope. Slope=$\frac{\Delta y}{\Delta x}$. So according to your hypothesis, you find the value of $\frac{\Delta x}{\Delta y}$. Take the inverse you get the value.