We apply the Chain rule to calculate
∂ G ∂ t = ∂ F ∂ z ∂ ( A t ) ∂ t = ∂ F ∂ z A ∂ G ∂ γ = ∂ F ∂ x ∂ ( γ + s ) ∂ γ + ∂ F ∂ y ∂ ( γ − s ) ∂ γ = ∂ F ∂ x + ∂ F ∂ y ∂ G ∂ s = ∂ F ∂ x ∂ ( γ + s ) ∂ s + ∂ F ∂ y ∂ ( γ − s ) ∂ s = ∂ F ∂ x − ∂ F ∂ y {\displaystyle \displaystyle {\begin{aligned}{\frac {\partial G}{\partial t}}&={\frac {\partial F}{\partial z}}{\frac {\partial (At)}{\partial t}}={\frac {\partial F}{\partial z}}A\\{\frac {\partial G}{\partial \gamma }}&={\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial \gamma }}={\frac {\partial F}{\partial x}}+{\frac {\partial F}{\partial y}}\\{\frac {\partial G}{\partial s}}&={\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial s}}+{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial s}}={\frac {\partial F}{\partial x}}-{\frac {\partial F}{\partial y}}\end{aligned}}}
For the second derivatives, we apply the Chain rule again to the already calculated derivatives
∂ 2 G ∂ γ 2 = ∂ ∂ x ∂ F ∂ x ∂ ( γ + s ) ∂ γ + ∂ ∂ y ∂ F ∂ x ∂ ( γ − s ) ∂ γ + ∂ ∂ x ∂ F ∂ y ∂ ( γ + s ) ∂ γ + ∂ ∂ y ∂ F ∂ y ∂ ( γ − s ) ∂ γ = ∂ 2 F ∂ x 2 + ∂ 2 F ∂ y ∂ x + ∂ 2 F ∂ x ∂ y + ∂ F 2 ∂ y 2 ∂ 2 G ∂ s 2 = ∂ ∂ x ∂ F ∂ x ∂ ( γ + s ) ∂ s + ∂ ∂ y ∂ F ∂ x ∂ ( γ − s ) ∂ s − ∂ ∂ x ∂ F ∂ y ∂ ( γ + s ) ∂ s − ∂ ∂ y ∂ F ∂ x ∂ ( γ − s ) ∂ s = ∂ 2 F ∂ x 2 − ∂ 2 F ∂ y ∂ x − ∂ 2 F ∂ x ∂ y + ∂ F 2 ∂ y 2 {\displaystyle \displaystyle {\begin{aligned}{\frac {\partial ^{2}G}{\partial \gamma ^{2}}}&={\frac {\partial }{\partial x}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma -s)}{\partial \gamma }}+{\frac {\partial }{\partial x}}{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma +s)}{\partial \gamma }}+{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma -s)}{\partial \gamma }}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\partial ^{2}F}{\partial x^{2}}}+{\frac {\partial ^{2}F}{\partial y\partial x}}+{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}\\{\frac {\partial ^{2}G}{\partial s^{2}}}&={\frac {\partial }{\partial x}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma +s)}{\partial s}}+{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma -s)}{\partial s}}-{\frac {\partial }{\partial x}}{\frac {\partial F}{\partial y}}{\frac {\partial (\gamma +s)}{\partial s}}-{\frac {\partial }{\partial y}}{\frac {\partial F}{\partial x}}{\frac {\partial (\gamma -s)}{\partial s}}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\partial ^{2}F}{\partial x^{2}}}-{\frac {\partial ^{2}F}{\partial y\partial x}}-{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}\end{aligned}}}
Now we set together
∂ 2 G ∂ γ 2 + ∂ G 2 ∂ s 2 = ∂ 2 F ∂ x 2 + ∂ 2 F ∂ y ∂ x + ∂ 2 F ∂ x ∂ y + ∂ F 2 ∂ y 2 + ∂ 2 F ∂ x 2 − ∂ 2 F ∂ y ∂ x − ∂ 2 F ∂ x ∂ y + ∂ F 2 ∂ y 2 = 2 ∂ 2 F ∂ x 2 + 2 ∂ 2 F ∂ y 2 = 2 ∂ F ∂ z {\displaystyle \displaystyle {\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial G^{2}}{\partial s^{2}}}={\frac {\partial ^{2}F}{\partial x^{2}}}+{\frac {\partial ^{2}F}{\partial y\partial x}}+{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}F}{\partial x^{2}}}-{\frac {\partial ^{2}F}{\partial y\partial x}}-{\frac {\partial ^{2}F}{\partial x\partial y}}+{\frac {\partial F^{2}}{\partial y^{2}}}=2{\frac {\partial ^{2}F}{\partial x^{2}}}+2{\frac {\partial ^{2}F}{\partial y^{2}}}=2{\frac {\partial F}{\partial z}}}
where the last step is already given by the problem.
With ∂ 2 G ∂ γ 2 + ∂ G 2 ∂ s 2 = 2 ∂ F ∂ z {\displaystyle \displaystyle {\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial G^{2}}{\partial s^{2}}}=2{\frac {\partial F}{\partial z}}} and ∂ G ∂ t = A ∂ F ∂ z {\displaystyle \displaystyle {\frac {\partial G}{\partial t}}=A{\frac {\partial F}{\partial z}}}
we conclude that A = 2 {\displaystyle \displaystyle A=2} is the value that ∂ 2 G ∂ γ 2 + ∂ G 2 ∂ s 2 = ∂ G ∂ t {\displaystyle \displaystyle {\frac {\partial ^{2}G}{\partial \gamma ^{2}}}+{\frac {\partial G^{2}}{\partial s^{2}}}={\frac {\partial G}{\partial t}}}
and 
is the value that 
