Solución
∂
z
∂
u
=
−
2
s
e
n
u
3
s
e
n
v
y
∂
z
∂
v
=
−
2
c
o
s
u
c
o
s
v
3
s
e
n
2
v
\frac{\partial z}{\partial u} = \frac{-2sen u}{3sen v}\text{ y } \frac{\partial z}{\partial v} = \frac{-2cos ucos v}{3sen^2v}
∂
u
∂
z
=
3
se
n
v
−
2
se
n
u
y
∂
v
∂
z
=
3
se
n
2
v
−
2
cos
u
cos
v