Partial Derivative of $xy^2+yz^2+xyz+x^2y^2z^2=5$ Someone can tell me what the Partial Derivative of $\frac{d^2z}{dy^2}$ of function $z(x,y)$ if it`s look like this:
$$xy^2+yz^2+xyz+x^2y^2z^2=5$$
I try to solve the first derivative:
$$\frac{dz}{dy}=(2yx+z^2z′+xzz′+2x^2yz^2z′)$$
but I am not sure if its okay,
Thank you!
 A: Putting the arguments in the given equality, this is
$$xy^2+yz(x,y)^2+xyz(x,y)+x^2y^2z(x,y)^2=5 .$$
Deriving with respect to $y$ gives
$$2xy + z(x,y)^2 + y 2z(x,y) \frac {\partial z} {\partial y}(x,y) + xz(x,y) + xy \frac {\partial z} {\partial y}(x,y) + x^2 2y z(x,y)^2 + x^2y^2 2 z(x,y) \frac {\partial z} {\partial y}(x,y) = 0 .$$
More nicely written,
$$2xy + z^2 + 2yz \frac {\partial z} {\partial y} + xz + xy \frac {\partial z} {\partial y} + 2x^2 y z^2 + 2x^2 y^2 z \frac {\partial z} {\partial y} = 0 ,$$
whence it follows that
$$\frac {\partial z} {\partial y} = -\frac {2xy + z^2 + xz + 2x^2 yz^2} {2yz + xy + 2x^2 y^2 z} .$$
Of course, one would have to make sure that the denominator exists, but since no domain of definition is given I shall ignore this issue and consider the computation to be purely formal. The expression obtained does not seem to be further simplifiable.
A: Hint: Note that by the product and the chain rule we have
$$
\frac{\partial}{\partial y} y^pz^q = py^{p-1}z^q + y^pqz^{q-1} \frac{\partial}{\partial y} z.
$$
