Answer
$${f_x}\left( {x,y} \right) = 2x - 2y{\text{ and }}{f_y}\left( {x,y} \right) = - 2x + 2y$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^2} - 2xy + {y^2} \cr
& {\text{Differentiate by using the limit definition}} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h,y} \right) - f\left( {x,y} \right)}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^2} - 2\left( {x + h} \right)y + {y^2} - \left( {{x^2} - 2xy + {y^2}} \right)}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} + 2xh + {h^2} - 2xy - 2yh - {x^2} + 2xy}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2xh + {h^2} - 2yh}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \left( {2x + h - 2y} \right) \cr
& {\text{Evaluate the limit}} \cr
& {f_x}\left( {x,y} \right) = 2x - 2y \cr
& \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x,y + h} \right) - f\left( {x,y} \right)}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - 2x\left( {y + h} \right) + {{\left( {y + h} \right)}^2} - \left( {{x^2} - 2xy + {y^2}} \right)}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2xy - 2xh + {y^2} + 2yh + {h^2} + 2xy - {y^2}}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2xh + 2yh + {h^2}}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \left( { - 2x + 2y + h} \right) \cr
& {\text{Evaluate the limit}} \cr
& {f_y}\left( {x,y} \right) = - 2x + 2y \cr} $$