Answer
$$\eqalign{
& f\left( {x,y} \right) = \frac{{1 - {x^2}}}{{{y^2}}} \cr
& dz = - 0.012037 \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{1 - {{\left( {3.05} \right)}^2}}}{{{{\left( {5.95} \right)}^2}}} - \frac{{1 - {3^2}}}{{{6^2}}} \cr
& {\text{Rewrite the expression}} \cr
& = \frac{{1 - {{\left( {3 + 0.05} \right)}^2}}}{{{{\left( {6 - 0.05} \right)}^2}}} - \frac{{1 - {3^2}}}{{{6^2}}} \cr
& z = \frac{{\overbrace {1 - {{\left( {3 + 0.05} \right)}^2}}^{1 - \left( {x + \Delta x} \right)}}}{{\underbrace {{{\left( {6 - 0.05} \right)}^2}}_{\left( {y + \Delta y} \right)}}} - \overbrace {\frac{{1 - {3^2}}}{{\underbrace {{6^2}}_{{y^2}}}}}^{1 - {x^2}} \cr
& {\text{Let the function }} \cr
& z = f\left( {x,y} \right) = \frac{{1 - {x^2}}}{{{y^2}}},{\text{ with }}x = 3{\text{ and }}y = 6,{\text{ }} \cr
& \Delta x = 0.05{\text{ and }}\Delta y = - 0.05 \cr
& {\text{Therefore, the total differential is }} \cr
& dz = \frac{\partial }{{\partial x}}\left[ {\frac{{1 - {x^2}}}{{{y^2}}}} \right]dx + \frac{\partial }{{\partial y}}\left[ {\frac{{1 - {x^2}}}{{{y^2}}}} \right]dy \cr
& dz = - \frac{{2x}}{{{y^2}}}dx + \left( { - \frac{{2\left( {1 - {x^2}} \right)}}{{{y^3}}}} \right)dy \cr
& dz = - \frac{{2x}}{{{y^2}}}dx - \frac{{2\left( {1 - {x^2}} \right)}}{{{y^3}}}dy \cr
& {\text{With }}dx = \Delta x{\text{ and }}dy = \Delta y \cr
& dz = - \frac{{2\left( 3 \right)}}{{{{\left( 6 \right)}^2}}}\left( {0.05} \right) - \frac{{2\left( {1 - {{\left( 3 \right)}^2}} \right)}}{{{{\left( 6 \right)}^3}}}\left( { - 0.05} \right) \cr
& dz = - 0.012037 \cr
& \cr
& f\left( {x,y} \right) = \frac{{1 - {x^2}}}{{{y^2}}} \cr
& dz = - 0.012037 \cr} $$
