f(x,y) = 2(x^2)y - 4x/y
Prove that f'y (with respect to y) = 2x^2 + 4x/(y^2) using the definition of a partial derivative (limit)
To take the derivative with respect to $\displaystyle y$, you treat $\displaystyle x$ as a constant and use the standard derivative rules.
$\displaystyle f(x, y) = 2x^2y - \frac{4x}{y}$
$\displaystyle = 2x^2y^1 - 4x\,y^{-1}$.
$\displaystyle \frac{\partial f}{\partial y} = 1\cdot 2x^2 y^{1-1} - 1(-4x)y^{-1-1}$
$\displaystyle = 2x^2y^0 + 4x\,y^{-2}$
$\displaystyle = 2x + \frac{4x}{y^2}$.
The calculation is tedious but straightforward.
$\displaystyle \frac{\partial f}{\partial y} = \lim_{h\to 0}\frac{f(x,y+h)-f(x,y)}{h}$
$\displaystyle =\lim_{h\to 0}\frac{2x^2(y+h)-\frac{4x}{y+h}-(2x^2y-\frac{4x}{y})}{h}$
$\displaystyle =\lim_{h\to 0}\frac{2hx^2-\frac{4x}{y+h}+\frac{4x}{y}}{h}$
$\displaystyle =\lim_{h\to 0}\bigg(2x^2+\frac{1}{h}\bigg(\frac{-4x(y)}{y(y+h)}+\frac{4x(y+h)}{y(y+h)}\bigg)\bigg)$
$\displaystyle =\lim_{h\to 0}\bigg(2x^2+\frac{-4xy+4xy+4xh}{h(y)(y+h)}\bigg)$
$\displaystyle =\lim_{h\to 0}\bigg(2x^2+\frac{4x}{(y)(y+h)}\bigg)$
$\displaystyle =2x^2+\frac{4x}{(y)(y+0)}$
$\displaystyle =2x^2+\frac{4x}{y^2}$
$\displaystyle \frac{f(x, y + h) - f(x, y)}{h}$
$\displaystyle = \frac{2x^2(y + h) - \frac{4x}{y + h} - \left(2x^2y - \frac{4x}{y}\right)}{h}$
$\displaystyle = \frac{2x^2h + \frac{4xh}{y(y + h)}}{h}$
$\displaystyle = \frac{\frac{2x^2h\,y(y + h) + 4h}{y(y + h)}}{h}$
$\displaystyle = \frac{2x^2h\,y(y + h) + 4xh}{h\,y(y + h)}$
$\displaystyle = \frac{h[2x^2y(y + h) + 4x]}{h\,y(y + h)}$
$\displaystyle = \frac{2x^2y(y + h) + 4x}{y(y + h)}$.
Now making $\displaystyle h \to 0$ we get
$\displaystyle \frac{2x^2y^2 + 4x}{y^2}$
$\displaystyle = 2x^2 + \frac{4x}{y^2}$.