Answer
$$\eqalign{
& \frac{{\partial f}}{{\partial x}} = y{e^{xy}} + \frac{{6x{e^{3{x^2}}}}}{{{e^{{y^2}}}}} \cr
& \frac{{{\partial ^2}f}}{{\partial x\partial y}} = {e^{xy}} + xy{e^{xy}} - 12xy{e^{3{x^2} - {y^2}}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {e^{xy}} + {e^{3{x^2} - {y^2}}} \cr
& {\text{Recall that }}{e^{a - b}} = \frac{{{e^a}}}{{{e^b}}} \cr
& f\left( {x,y} \right) = {e^{xy}} + \frac{{{e^{3{x^2}}}}}{{{e^{{y^2}}}}} \cr
& {\text{Find }}{f_x} \cr
& {f_x} = \frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{e^{xy}} + \frac{{{e^{3{x^2}}}}}{{{e^{{y^2}}}}}} \right] \cr
& \frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{e^{xy}}} \right] + \frac{1}{{{e^{{y^2}}}}}\frac{\partial }{{\partial x}}\left[ {{e^{3{x^2}}}} \right] \cr
& {\text{Consider }}y{\text{ as a constant}} \cr
& \frac{{\partial f}}{{\partial x}} = y{e^{xy}} + \frac{1}{{{e^{{y^2}}}}}\left( {6x{e^{3{x^2}}}} \right) \cr
& \frac{{\partial f}}{{\partial x}} = y{e^{xy}} + \frac{{6x{e^{3{x^2}}}}}{{{e^{{y^2}}}}} \cr
& \cr
& {\text{Find }}{f_{xy}} \cr
& {f_{xy}} = \frac{{{\partial ^2}f}}{{\partial x\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{{\partial f}}{{\partial x}}} \right] \cr
& \frac{{{\partial ^2}f}}{{\partial x\partial y}} = \frac{\partial }{{\partial y}}\left[ {y{e^{xy}} + \frac{{6x{e^{3{x^2}}}}}{{{e^{{y^2}}}}}} \right] \cr
& \frac{{{\partial ^2}f}}{{\partial x\partial y}} = \frac{\partial }{{\partial y}}\left[ {y{e^{xy}}} \right] + \frac{\partial }{{\partial y}}\left[ {\frac{{6x{e^{3{x^2}}}}}{{{e^{{y^2}}}}}} \right] \cr
& \frac{{{\partial ^2}f}}{{\partial x\partial y}} = \frac{\partial }{{\partial y}}\left[ {y{e^{xy}}} \right] + 6x{e^{3{x^2}}}\frac{\partial }{{\partial y}}\left[ {{e^{ - {y^2}}}} \right] \cr
& {\text{Using the product rule}} \cr
& \frac{{{\partial ^2}f}}{{\partial x\partial y}} = {e^{xy}}\frac{\partial }{{\partial y}}\left[ y \right] + y\frac{\partial }{{\partial y}}\left[ {{e^{xy}}} \right] + 6x{e^{3{x^2}}}\frac{\partial }{{\partial y}}\left[ {{e^{ - {y^2}}}} \right] \cr
& \frac{{{\partial ^2}f}}{{\partial x\partial y}} = {e^{xy}} + y\left( {x{e^{xy}}} \right) + 6x{e^{3{x^2}}}\left( { - 2y{e^{ - {y^2}}}} \right) \cr
& \frac{{{\partial ^2}f}}{{\partial x\partial y}} = {e^{xy}} + xy{e^{xy}} - 12xy{e^{3{x^2} - {y^2}}} \cr} $$
