Finding $\frac{\partial^6 f}{\partial x^4 \partial y^2}$ 
Find $$\frac{\partial^6 f}{\partial x^4 \partial y^2}(0,0)$$ Of the $f(x,y)=\frac{1}{1-x^2y}$

$$\frac{1}{1-x^2y}=\sum_{n=1}^{\infty}(x^2y)^n$$ as $|x^2y|<1$
So we have $\frac{1}{1-x^2y}\approx1+x^2y+x^4y^2$
But how should we continue from here?
 A: Why don't you just derivate? First with respect to $y$ two times for example, to get $2x^{4}+6x^{6}y$, and then after derivating four times with respect to $x$ and evaluate at $0$ the only term you will get is $2\cdot 4!=48$.
A: Your method is correct:
$$z=\sum_{n=\color{red}{0}}^{\infty} (x^2y)^n=1+x^2y+x^4y^2+x^6y^3+\cdots$$
$$z_{x^4}=4!y^2+4!x^2y^3+\cdots$$
$$z_{x^4y^2}=4!\cdot 2!+4!x^2\cdot 2!y+\cdots$$
$$z_{x^4y^2}(0,0)=4!2!=48.$$
A: Use the below statement before starts to differentiate.
$$\frac{\partial^6 f(x,y)}{\partial x^4 \partial y^2}=\frac{\partial^4 \left(\frac{\partial^2 f(x,y)}{\partial y^2}\right)}{\partial x^4}$$
Calculate $\displaystyle \frac{\partial^2 f(x,y)}{\partial y^2}$ for first.
$$\displaystyle \frac{\partial f(x,y)}{\partial y}=\frac{x^2}{(1 - x^2 y)^2}$$
$$\displaystyle \frac{\partial^2 f(x,y)}{\partial y^2}=\frac{(2 x^4)}{(1 - x^2 y)^3}$$
Now you should substitute $y=0$ since you are not required to differentiate by $y$ it takes place only as a constant while you differentiate by $x$.
$$\displaystyle \frac{\partial^2 f(x,0)}{\partial y^2}=\frac{(2 x^4)}{(1 - x^2\cdot  0)^3}=2x^4$$
It is now much easier to differentiate by $x$ four times.
$$\displaystyle \frac{\partial^4 (2x^4)}{\partial x^4}=\frac{\partial^3 (4\cdot 2 x^3)}{\partial x^3}=\frac{\partial^2 (4\cdot 3 \cdot 2 x^2)}{\partial x^2}=\frac{\partial^2 (4\cdot 3 \cdot 2 \cdot 2 x)}{\partial x}=4\cdot 3 \cdot 2 \cdot 2=48.$$
