Implicit derivation of of $x^y+\sin y=1$ at point $(1,0)$. So first I show that $y$ is dependent on $x$. Let $f(x,y)=x^y+\sin y$ then $f_y=x^y\ln x+\cos y$ and $f_y(1,0)=1\neq0$.
So if we derive $f$ with respect to $x$ we get $x^yy'\ln x+y'\cos y=0\iff y'(x)=0$ but $y'(x)=\frac{-f_x}{f_y}=\frac{-yx^{y-1}}{x^y\ln x+\cos y}$.
I should always be able to use the formula and directly derive and get the same answer right? So what is wrong here?
 A: First - Gradient related to the normal surface
If you want to impose a point for derivative you have to consider a surface as follow
the surface can be defined as :
$$ x^y+\sin(y)-1 =0$$
Then defining the function
$$ f : (x,y) \to x^y+\sin(y)-1$$
The gradient can be calculated as
$$ \nabla f(x,y)=[yx^{y-1} \  \ln(x)x^y +\sin(y)]$$
At $(1,0)$ it gives :
$$ \nabla f (1,0)=[0   \ 0 ]$$
Second - Implicit derivation
Starting back from
$$ x^y + \sin(y) - 1= 0$$
Noting $ y ( \cdot x )$.
Gives by replacing
$$ x^{y(x)}+\sin(y(x))-1=0$$
Derivating wrt $x$ :
$$ (y'(x)\ln(x)+y(x)/x)+y'(x)\cos(x)=0 $$
Gives
$$ y'(x) = -\dfrac{y(x)}{x(\ln(x)+\cos(x))} $$
You can get $y(0)$ from the very first equation.
A: For clarity, we better consider the equation as
$$x^{y(x)}+\sin (y(x))=1$$
Then $$
\frac{d}{d x}\left(x^{y(x)}\right)+ \frac{d}{d x}\sin (y(x))=0
$$
By logarithmic differentiation, we have
$$
\frac{d}{d x}\left(x^{y(x)}\right)\Big|_{(1,0)}=x^{y(x)}\left(\frac{y(x)}{x}+\frac{d y(x)}{d x}\cdot \ln x\right) \Big|_{(1,0)}=0
$$
and
$$
\begin{aligned}
\left.\frac{d}{d x} \sin (y(x))\right|_{(1,0)} & =\left.\cos (y(x)) \frac{d y}{d x}\right|_{(1,0)} \\
& =\left.\frac{d y}{d x}\right|_{(1,0)}
\end{aligned}
$$
We can now conclude that $$
0+\left.\frac{d y}{d x}\right|_{(1,0)}=\left.0 \Rightarrow \frac{d y}{d x}\right|_{(1,0)}=0
$$
