Answer
a) $\dfrac{1}{x^2}=y^2$
b) $\dfrac{1}{(x+3)^2}=y^2$
c) $\dfrac{1}{(x+C)^2}=y^2$
Work Step by Step
a) $y^2=\dfrac{1}{x^2}$
Thus, $y'=-[(-1) x^{-2}]=\dfrac{1}{x^2}=y^2$
b) $y^2=\dfrac{1}{(x+3)^2}$
Thus, $y'=-[(-1) (x+3)^{-2}]=\dfrac{1}{(x+3)^2}=y^2$
c) $y^2=\dfrac{1}{(x+C)^2}$
Thus, $y'=-[(-1) (x+C)^{-2}]=\dfrac{1}{(x+C)^2}=y^2$