Hello,
$\displaystyle \boxed{e^{xy} \sin(yz)-e^{yz}\cos(xz)+z=0}$
--> $\displaystyle ?=\frac{\partial z}{\partial y}=z'(y)$ (for more comfort)
---> $\displaystyle \boxed{e^{x{\color{red}y}} \sin({\color{red}y}{\color{blue}z})-e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu e}z})+{\color{blue}z}=0}$
~~~~~~~~~~
The first term will use the product rule since there are y and z in it, considering x as a constant and y as the variable.
Thus the derivative of the first term will be :
$\displaystyle xe^{xy} \sin(yz)+e^{xy} {\color{red}\dots} \cos(xy)$
The red dots correspond to the derivative of yz. We have to use the product rule.
---> $\displaystyle {\color{red}\dots}=z+yz'$
---------> $\displaystyle \boxed{xe^{xy} \sin(yz)+e^{xy} (z+yz') \cos(xy)} \ (1)$
~~~~~~~~~~
Now considering the second term $\displaystyle e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu e}z})$
We have to use the product rule too.
$\displaystyle \underbrace{\color{red}\dots}_{\text{derivative of yz, again}} e^{yz} \cos(xz) \ \ + \ \ e^{yz} (-(xz') \sin(xz))$
---------> $\displaystyle \boxed{(z+yz') e^{yz} \cos(xz) - xz' e^{yz} \sin(xz)} \ (2)$
~~~~~~~~~~
Now, the third term, z.
Its derivative is z' (3)
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The derivative of the whole thing will be $\displaystyle (1)-(2)+(3)=0$
And... I won't copy it, you can do it
I hope there is no mistake, check it again, it will make a good exercise