Hello, Melcarthus!
Your canceling is incorrect.
$\displaystyle 35.\;\begin{Bmatrix}xy^2 &=& 10^{11} & [1] \\ \\[-3mm] \dfrac{x^3}{y} &=& 10^{12} & [2]\end{Bmatrix}$
Divide [2] by [1]: .$\displaystyle \dfrac{\frac{x^3}{y}}{xy^2} \:=\:\frac{10^{12}}{10^{11}} \quad\Rightarrow\quad \frac{x^3}{xy^3} \:=\:10 \quad\Rightarrow\quad x^2 \:=\:10y^3 \;\;[3]$
Square [1]: .$\displaystyle (xy^2)^2 \:=\:(10^{11})^2 \quad\Rightarrow\quad x^2y^4 \:=\:10^{22}\;\;[4]$
Substitute [3] into [4]: .$\displaystyle (10y^3)(y^4) \:=\:10^{22} \quad\Rightarrow\quad y^7 \:=\:10^{21} \quad\Rightarrow\quad \boxed{y \:=\:10^3}$
Substitute into [1]: .$\displaystyle x(10^3)^2 \:=\:10^{11} \quad\Rightarrow\quad x(10^6) \:=\:10^{11} \quad\Rightarrow\quad \boxed{x \:=\:10^5}$