Use implicit differentiation to determine $\frac{\partial z}{\partial x}$ in $yz=ln(x+z)$ and $ \sin(xyz)=x+2y+3z$.
Here is my answer:
$$ yz=ln(x+z) $$ $$ yz'=(1+z')\frac{1}{x+z} $$ $$ z' = \frac{1}{yx+yz-1} $$
and
$$ \sin(xyz)=x+2y+3z $$ $$ y(z+xz')\cos(xyz)=1+3z' $$ $$ yz\cos(xyz)+xyz'\cos(xyz) = 1+3z' $$ $$ yz\cos(xyz) - 1 = z'(3-xy\cos(xyz)) $$ $$ z' = \frac{yz\cos(xyz) - 1}{3-xy\cos(xyz)} $$
is that right?
Not quite. Don't forget the chain rule, and to include other instances of $\frac{\partial y}{\partial x}$.
For instance, in the first, don't forget that $$ \frac{\partial}{\partial x}(yz) = \frac{\partial y}{\partial x}z + y\frac{\partial z}{\partial x} $$
and in the second, don't forget that $$ \frac{\partial}{\partial x} ( x + 2y + 3z) = 1 + 2\frac{\partial y} {\partial x} + 3\frac{\partial z}{\partial x} $$
Otherwise, your techniques look reasonably ok.