I am new to calculus and am somewhat confused by the details of partial derivatives in more complex constructions. I was asked to work out the following and was wondering if anybody could give me some feedback. Many thanks in advance!
The function $Z(x,y)$ is defined by the formula $$yz + \sin(z) + xyz - y\cos (z) = -1$$ around point $(1,1,0)$. Calculate $\frac{\partial Z}{\partial y}(1,1)$.
Here is what I've tried:
$$\frac{\partial Z}{\partial y} = z + yz_{y} + \cos(z)·z_{y}+x·z_{y} - \cos(z)+y·\sin(z)·z_{y}=0$$
$$z_{y}(y+\cos(z)+x+y·\sin(z)) = -z+\cos(z)$$
$$z_{y} = \frac{-z + \cos(z)}{y + \cos(z)+ x +y·\sin(z)}$$
By putting in $(1,1,0)$, I then get to $$\frac{\partial Z}{\partial y}(1,1)=\frac{1}{3}$$
$$yz+\sin(z)+xyz-y\cos(z)=-1$$
Taking partial derivative with respect to $y$, $$ z + yz_{y} + \cos(z)z_{y}+x z +xyz_y - \cos(z)+y·\sin(z)·z_{y}=0$$
Now, substitute $(1,1,0)$,
$$0+z_y+\cos (0)z_y+0+(1)(1)z_y-\cos(0)+\sin(0)z_y=0$$
$$3z_y-1=0$$
$$z_y=\frac13$$