Find the tangent and normal bundle of $Y$ where the mapping $f:[-1,1]\times[1,2]\to Y$ is defined as
$$f(x,y)=\left(x\sqrt{\frac{y}{1+x^2}},\sqrt{\frac{y}{1+x^2}},xy\right).$$ Then show that $Y$ is orientable.
So as far as I know, the tangent bundle is determined by $f$ and a linear combination of $f_x$ and $f_y$. So I calculated both partial derivatives and so the tangent bundle would be $(f(x,y),Lin\{f_x(x,y),f_y(x,y)\})$ where $(x,y)\in[-1,1]\times[1,2]$. Is this correct?
If yes, then the normal bundle would be $(f(x,y),Lin\{f_x(x,y)\times f_y(x,y)\})$ so we take the cross product of the partials. I also am not sure how to prove that we can orient $Y$ and how to choose the orientation. Any suggestions?
Your proof for the tangent and normal bundle are correct. For the orientation you want to used the induced orientation from $[-1,1]\times [1,2]$. Then you need to check whether $f$ is injective. If it is you are done, if not you need to check whether distinct points in $[-1,1]\times [1,2]$ that map to the same point have the same orientation. Essentially you want to distinguish whether the image of $f$ is just a deformed rectangle, some flat ribbon or a version of the Möbius strip.
Edit: I claim that $f$ is injective on the given domain. Suppose $f(x_1,y_1)=f(x_2,y_2)$, then $$x_1\sqrt{\frac{y_1}{1+x_1^2}}=x_2\sqrt{\frac{y_2}{1+x_2^2}}$$ and $$\sqrt{\frac{y_1}{1+x_1^2}}=\sqrt{\frac{y_2}{1+x_2^2}}$$ These can only both be satisfied if either $\sqrt{\frac{y_1}{1+x_1^2}}=0$ which can't happen for $y \in[1,2]$ or if $x_1=x_2$. If $x_1=x_2$, the third equation $$x_1y_1=x_2y_2$$ implies $y_1=y_2$ or $x_1=0$ and $f(0,y)$ is injective as well.