I have tried to calculate this question and am not sure about my process being correct. If you have the time, can you tell me if my answer is correct or not and if not where I have made my mistake(s). Thanks in advance.
$$ \begin{align*} f &= f(x.y), x = u^2 - v^2, y = \frac{u}{v}\\ &\text{find $f_{uu}$ and $f_{vv}$}\\ \frac{\partial x}{\partial u} &= 2u\\ \frac{\partial x}{\partial v} &= -2v\\ \frac{\partial y}{\partial u} &= \frac{1}{v}\\ \frac{\partial y}{\partial v} &= -\frac{u}{v^2}\\ \frac{\partial f}{\partial u} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}\\ &=2u\left(\frac{\partial f}{\partial x}\right) + \frac{1}{v}\left(\frac{\partial f}{\partial y}\right)\\ \frac{\partial}{\partial u}\left(f\right) &= 2u\frac{\partial}{\partial x}\left( f \right) + \frac{1}{v}\frac{\partial}{\partial y}\left( f \right) \tag{1} \label{equation0}\\ \frac{\partial^2 f}{\partial u^2} &= \left(\frac{\partial}{\partial u}\right) \left(\frac{\partial f}{\partial u}\right)\\ &=2\left(\frac{\partial f}{\partial x}\right) + 2u\left(\frac{\partial}{\partial u}\right)\left(\frac{\partial f}{\partial x}\right) + \frac{1}{v}\left(\frac{\partial}{\partial u}\right)\left(\frac{\partial f}{\partial y}\right) \tag{2} \label {equation1}\\ &\text{Use equation $\eqref{equation0}$ to find $\frac{\partial}{\partial u} \left(\frac{\partial f}{\partial x} \right)$ and $\frac{\partial}{\partial u} \left(\frac{\partial f}{\partial y} \right)$.}\\ \frac{\partial}{\partial u} \left(\frac{\partial f}{\partial x}\right) &= 2u\left(\frac{\partial}{\partial x}\right) \left(\frac{\partial f}{\partial x}\right) + \frac{1}{v} \left(\frac{\partial}{\partial y}\right) \left(\frac{\partial f}{\partial x}\right) \\ &= 2u\frac{\partial^2 f}{\partial x^2} + \frac{1}{v} \frac{\partial^2 f}{\partial y \partial x} \tag{3} \label{equation2} \\ \frac{\partial}{\partial u} \left(\frac{\partial f}{\partial y}\right) &= 2u \left(\frac{\partial}{\partial x}\right) \left(\frac{\partial f}{\partial y}\right) + \frac{1}{v} \left(\frac{\partial}{\partial y}\right) \left(\frac{\partial f}{\partial y}\right) \\ &= 2u \frac{\partial^2 f}{\partial x \partial y} + \frac{1}{v} \frac{\partial^2 f}{\partial y^2} \tag{4} \label{equation3} \\ &\text{Substitute $\eqref{equation2}$ and $\eqref{equation3}$ into $\eqref{equation1}$.}\\ \frac{\partial^2 f}{\partial u^2} &= 2 \left(\frac{\partial f}{\partial x}\right) + 2u \left(2u \frac{\partial^2 f}{\partial x^2} + \frac{1}{v} \frac{\partial^2 f}{\partial y \partial x}\right) + \frac{1}{v} \left(2u \frac{\partial^2 f}{\partial x \partial y} + \frac{1}{v} \frac{\partial^2 f}{\partial y^2}\right)\\ &= 2 \frac{\partial f}{\partial x} + 4u^2 \frac{\partial^2 f}{\partial x^2} + \frac{2u}{v} \frac{\partial^2 f}{\partial y \partial x} + \frac{2u}{v} \frac{\partial^2 f}{\partial x \partial y} + \frac{1}{v^2} \frac{\partial^2 f}{\partial y^2}\\ &\text{Assuming $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$}\\ &= 2 \frac{\partial f}{\partial x} + 4u^2 \frac{\partial^2 f}{\partial x^2} + \frac{4u}{v} \left( \frac{\partial^2 f}{\partial y \partial x} \right) + \frac{1}{v^2} \left(\frac{\partial^2 f}{\partial y^2}\right)\\ \frac{\partial f}{\partial v} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial v}\\ &=-2v\left(\frac{\partial f}{\partial x}\right) - \frac{u}{v^2} \left(\frac{\partial f}{\partial y}\right)\\ \frac{\partial}{\partial v}\left(f\right) &= -2v \frac{\partial}{\partial x}\left( f \right) - \frac{u}{v^2}\frac{\partial}{\partial y}\left( f \right) \tag{5} \label{equation4}\\ \frac{\partial^2 f}{\partial v^2} &= \left(\frac{\partial}{\partial v}\right) \left(\frac{\partial f}{\partial v}\right)\\ &=-2\left(\frac{\partial f}{\partial x}\right) - 2v \left(\frac{\partial}{\partial v}\right)\left(\frac{\partial f}{\partial x}\right) + \frac{2u}{v^3}\left(\frac{\partial f}{\partial y}\right) - \\ & \frac{u}{v^2} \left(\frac{\partial}{\partial v}\right)\left(\frac{\partial f}{\partial y}\right) \tag{6} \label {equation5}\\ &\text{Use equation $\eqref{equation4}$ to find $\frac{\partial}{\partial v} \left(\frac{\partial f}{\partial x} \right)$ and $\frac{\partial}{\partial v} \left(\frac{\partial f}{\partial y} \right)$.}\\ \frac{\partial}{\partial v} \left(\frac{\partial f}{\partial x}\right) &= -2v \left(\frac{\partial}{\partial x}\right) \left(\frac{\partial f}{\partial x}\right) - \frac{u}{v^2} \left(\frac{\partial}{\partial y}\right) \left(\frac{\partial f}{\partial x}\right) \\ &= -2v \frac{\partial^2 f}{\partial x^2} - \frac{u}{v^2} \frac{\partial^2 f}{\partial y \partial x} \tag{7} \label{equation6} \\ \frac{\partial}{\partial v} \left(\frac{\partial f}{\partial y}\right) &= -2v \left(\frac{\partial}{\partial x}\right) \left(\frac{\partial f}{\partial y}\right) - \frac{u}{v^2} \left(\frac{\partial}{\partial y}\right) \left(\frac{\partial f}{\partial y}\right) \\ &= -2v \frac{\partial^2 f}{\partial x \partial y} - \frac{u}{v^2} \frac{\partial^2 f}{\partial y^2} \tag{8} \label{equation7} \\ &\text{Substitute $\eqref{equation6}$ and $\eqref{equation7}$ into $\eqref{equation5}$.}\\ \frac{\partial^2 f}{\partial v^2} &= -2 \left(\frac{\partial f}{\partial x}\right) - 2v \left(-2v \frac{\partial^2 f}{\partial x^2} - \frac{u}{v^2} \frac{\partial^2 f}{\partial y \partial x}\right) + \frac{2u}{v^3} \left(\frac{\partial f}{\partial y}\right) \\ &- \frac{u}{v^2} \left(-2v \frac{\partial^2 f}{\partial x \partial y} - \frac{u}{v^2} \frac{\partial^2 f}{\partial y^2}\right)\\ &= -2 \frac{\partial f}{\partial x} + 4v^2 \frac{\partial^2 f}{\partial x^2} + \frac{2u}{v} \frac{\partial^2 f}{\partial y \partial x} - \frac{2u}{v^3} \frac{\partial f}{\partial y} + \frac{2u}{v} \frac{\partial^2 f}{\partial x \partial y} \\ &+ \frac{u^2}{v^4} \frac{\partial^2 f}{\partial y^2}\\ &\text{Assuming $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$}\\ &= -2 \frac{\partial f}{\partial x} + 4v^2 \frac{\partial^2 f}{\partial x^2} - \frac{4u}{v} \left(\frac{\partial^2 f}{\partial y \partial x}\right) +\frac{2u}{v^3} \left(\frac{\partial f}{\partial y}\right) + \frac{u^2}{v^4} \left(\frac{\partial^2 f}{\partial y^2}\right)\\ \end{align*} $$
It is faster to use subscript notation. $$f_u=f_x u_x+f_y y_u=2uf_x+\frac{1}{v}f_y$$ Taking the derivative of the last expression again wrt. u, $$f_{uu}=2f_x+2u(2uf_{xx}+\frac{1}{v}f_{xy})+\frac{1}{v}(2uf_{xy}+\frac{1}{v}f_{yy})$$ $$f_{uu}=2f_x+4u^2f_{xx}+\frac{4u}{v}f_{xy}+\frac{1}{v^2}f_{yy}$$ You did it correct. Similarly, $$f_v=f_x x_v+f_y y_v=-2vf_x-\frac{u}{v^2}f_y$$ Taking the derivative of the last expression again wrt. v, $$f_{vv}=-2f_x-2v(f_{xx}x_v+f_{xy}y_v)+\frac{2u}{v^3}f_y-\frac{u}{v^2}(f_{xy}x_v+f_{yy}y_v)$$ $$f_{vv}=-2f_x+\frac{2u}{v^3}f_y-2v(f_{xx}(-2v)+f_{xy}(-\frac{u}{v^2}))-\frac{u}{v^2}(f_{xy}(-2v)+f_{yy}(-\frac{u}{v^2}))$$ $$f_{vv}=-2f_x+\frac{2u}{v^3}f_y+4v^2f_{xx}+\frac{4u}{v}f_{xy}+\frac{u^2}{v^4}f_{yy}$$ One term is different. Yours is $-\frac{4u}{v}f_{xy}$, mine is $+\frac{4u}{v}f_{xy}$. I guess mine is correct. But, we can never be %100 sure in math.