$X$ and $Y$ are independent random variables with standard normal distributions. Let $Z=X^2+Y$. Find $E(Z/X)$.
$Z/X$ is $X+Y/X$. $E(Z/X) = E(X)+E(Y/X)$. Now can we say that $Y/X$ follows the $t_1$ distribution and hence the expectation is $0$. Or can we use the fact that $X$ and $Y$ are independent and identical to arrive at an answer without going into the distribution of $Y/X$?
My answer addresses (what I believe to be) the question you meant to ask, and not the question as literally stated since there are mistakes in the question itself that make the answer uninteresting.
You asked if $\mathbb E(Y/X)$ is always $0$ without doing any integrals. The wiki page on the Cauchy distribution states two things of concern to us: first, that it is the ratio of two independent standard normal variables, and second, that its mean is undefined.
However, if you consider the same question but with another distribution in place of the normal distribution, you can avoid this technicality. Thus, I will modify the question instead of having $X,Y$ be standard normal, to instead assume the following:
Under these assumptions, we can compute $\mathbb E(Y/X)$ using conditional expectation.
$$ \mathbb E\left(\frac{Y}{X}\mid X\right)=\frac{\mathbb EY}{X}=0,\quad \text{almost surely}. $$ Thus, by the law of total probability $$ \mathbb E\frac{Y}{X}=\mathbb E\left(\frac{\mathbb EY}{X}\right)=0. $$
