Find $f'(c)$ using the derivative definition for $f(x)=\frac{1}{x^2}$ So $f:\mathbb R \backslash\{0\}\rightarrow\mathbb R$ by $f(x)=\frac{1}{x^2}$
I'm need to use the definition of the derivative to find $f'(c)$ for $\frac{1}{x^2}$.  If I use standard differentiation techniques I get:
$-\frac{2}{c^3}$
However, using the following definition:
$f'(c) = \frac{f(x)-f(c)}{x-c}$ 
I end up with $\frac{c^2-x^2}{x^2c^2(x-c)}$
What am I doing wrong?

set $x-c=h$ then $$f^{ \prime  }\left( c \right) =\lim _{ h\rightarrow 0 }{ \frac { f\left( c+h \right) -f\left( c \right)  }{ h }  } =\lim _{ h\rightarrow 0 }{ \frac { \frac { 1 }{ { \left( c+h \right)  }^{ 2 } } -\frac { 1 }{ { c }^{ 2 } }  }{ h }  } =$$ $$ \\ \\ =\lim _{ h\rightarrow 0 }{ \frac { { c }^{ 2 }-{ \left( c+h \right)  }^{ 2 } }{ { \left( c+h \right)  }^{ 2 }{ c }^{ 2 }h }  } =\lim _{ h\rightarrow 0 }{ \frac { -h\left( 2c+h \right)  }{ { \left( c+h \right)  }^{ 2 }{ c }^{ 2 }h }  } =\frac { -2 }{ { c }^{ 3 } } $$
