Let $f$ be a function of $x$ and $y$, where $x=u+v-uv$ and $y=uv-u$. Find $\dfrac{\partial ^2f}{\partial x^2}$in terms of $\dfrac{\partial ^2f}{\partial u^2}$, $\dfrac{\partial ^2f}{\partial u\partial v}$ and $\dfrac{\partial ^2f}{\partial v^2}$.
I got $$\dfrac{\partial ^2f}{\partial u^2}=f_{xu}(1-v)+f_{yu}(v-1)$$ $$\dfrac{\partial ^2f}{\partial v^2}= f_{xv}(1-u)+f_{yv}u$$ $$\dfrac{\partial ^2f}{\partial u\partial v}=f_{xu}(1-u)+f_x(-1)+f_{yu}u+f_y$$
How do I get $\dfrac{\partial ^2f}{\partial x^2}$ as there is no such term generated?
I think you have done the derivatives incorrectly. For example, the first one should be
\begin{align} \frac{\partial^2f}{\partial u^2} & = \frac{\partial}{\partial u}\bigg(\frac{\partial f}{\partial x}[1-v] + \frac{\partial f}{\partial y}[v-1] \bigg) \\ & = \bigg(\frac{\partial^2 f}{\partial x^2}(1-v) + \frac{\partial^2 f}{\partial x\partial y}(v-1) \bigg)[1-v] + \bigg(\frac{\partial^2 f}{\partial x\partial y}(1-v) + \frac{\partial^2 f}{\partial y^2}(v-1)\bigg)[v-1] \\ & = (1-v)^2\bigg(\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} + \frac{\partial^2 f}{\partial y^2} \bigg) \end{align}
I leave the rest to you.
EDIT:
Apologies. What you have done is not incorrect. You just need to apply the chain rule once more.
For example, you have
$$\dfrac{\partial ^2f}{\partial u^2}=f_{xu}(1-v)+f_{yu}(v-1)$$
but the $f_{xu}$ can be further decomposed as
$$f_{xu} = [f_x]_u = [f_x]_xx_u + [f_x]_yy_u$$