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The “slope” is what is sought (1st deriv.) not ‘x’ nor’y.’
Since you already have the equation that represents the function, neither of the variables is an unknown.
It is amazing how many people can’t focus on what they are really doing. Even “common core” may be above their heads.
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Can you not get it that f(x) means some function of x. As in say f(x) = x^2 (x squared) and then f'=2x where f' is the notation for denoting the first derivative of f(x) which in this case is x squared. Now just because f(x) can be represented as y (as in y=x^2) does not make that a multivariate differential calculus problem! Find f'(x,y) where f(x,y)=(y^2)(x^2) would be a multivariate differential calculus problem.
I figured you where to dense to simply admit your mistake.