Implicit function theorem usage Suppose I have the following equation representing level curves of a function:

$$-\left(\frac{1-\alpha}{x}+\frac{\alpha}{y}\right)=k \  \text{where} \ 0&lt;\alpha&lt;1$$

And I want to find $\frac{dy}{dx}$:

Using the implicit function theorem, I get:

$$\frac{dy}{dx}=-\frac{1-\alpha}{\alpha}\left(\frac{y}{x}\right)^2$$

However, I can also put the first equation into a $y$ in function of $x$ form:

$$y=-\frac{\alpha x}{(1-\alpha)+kx}$$

Which gives me:

$$\frac{dy}{dx}=-\frac{\alpha(1-\alpha)}{[(1-\alpha)+kx]^2}$$

I don't understand why the results are different. In the first expression, the constant $k$ disappears, while in the second equation it doesn't. Also, we have $y$ appearing in the first expression but obvisouly not in the second. How can I explain this?
Making use of fredgoodman's comment, we substitute $y$ into the first $\frac{dy}{dx}$ expression and get the second one.
Making use of fredgoodman's comment, we substitute $y$ into the first $\frac{dy}{dx}$ expression and get the second one.