Is it always possible to isolate $\frac {dy}{dx}$ in an implicit equation: $f(x,y)=k$?

Question

Is it always possible to isolate $\frac {dy}{dx}$ in an implicit equation: $f(x,y)=k$?

For example: $x^y+\ln(y)=3 \rightarrow y'+\frac y{x \ln x}+\frac {y'}y=0 \rightarrow y'=-\frac y{x \ln x}\cdot \frac 1 {1+\frac 1 y}$

Answer 1

Yes: this is clear when you write it in terms of differentials: $$ 0 = df = \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy $$ Then by definition $$ 0 = \frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}, $$ and clearly if $\partial f/\partial y \neq 0$ you can solve for $dy/dx$.

Answer 2

Yes, assuming that $f(x,y)$ actually has a non-trivial dependence on y. Just differentiate both sides with respect to $x$ to get $f_x+f_yy'=0$ and solve for $y'$.