1. **State the problem:** Find the derivative of the function $f(x) = \frac{1}{x^2}$ using the first principle of derivatives. 2. **Recall the definition of derivative from first principles:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the function to the formula:** $$f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$$ 4. **Find a common denominator for the numerator:** $$\frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{x^2 - (x+h)^2}{x^2 (x+h)^2}$$ 5. **Substitute back into the limit:** $$f'(x) = \lim_{h \to 0} \frac{\frac{x^2 - (x+h)^2}{x^2 (x+h)^2}}{h} = \lim_{h \to 0} \frac{x^2 - (x+h)^2}{h \cdot x^2 (x+h)^2}$$ 6. **Expand the squared term:** $$(x+h)^2 = x^2 + 2xh + h^2$$ 7. **Simplify the numerator:** $$x^2 - (x^2 + 2xh + h^2) = x^2 - x^2 - 2xh - h^2 = -2xh - h^2$$ 8. **Rewrite the limit:** $$f'(x) = \lim_{h \to 0} \frac{-2xh - h^2}{h \cdot x^2 (x+h)^2}$$ 9. **Factor out $h$ in the numerator:** $$f'(x) = \lim_{h \to 0} \frac{h(-2x - h)}{h \cdot x^2 (x+h)^2}$$ 10. **Cancel $h$ in numerator and denominator:** $$f'(x) = \lim_{h \to 0} \frac{\cancel{h}(-2x - h)}{\cancel{h} \cdot x^2 (x+h)^2} = \lim_{h \to 0} \frac{-2x - h}{x^2 (x+h)^2}$$ 11. **Evaluate the limit as $h \to 0$:** $$f'(x) = \frac{-2x}{x^2 \cdot x^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}$$ **Final answer:** $$f'(x) = -\frac{2}{x^3}$$