Bound the gradient of a smooth function $\nabla_y \Omega(y/|y|)$

Question

I am reading a book about the Euler equation and in order to prove a stability result, study a property of a Calderon-Zygmund operator and at some point the author uses something I do not understand. We consider $\Omega : \mathbb S^{d-1} \to \mathbb R$ a smooth function such that it satisfies a cancelation property, i.e. $$\int_{\mathbb S^{d-1}} \Omega (z) dz = 0.$$ At some point the author claim that we have $$|\nabla_y \Omega (y/|y|)| \le C \frac{1}{|y|}$$ for $y \in \mathbb R^d$ and $y \neq 0$. I do not really get this result because if we look at it componentwise, we get \begin{align} \bigg|\frac{\partial}{\partial y_i} \Omega \left(\frac{y}{|y|}\right)\bigg|&\le \sum_j\bigg|\frac{\partial \Omega}{\partial z_j}\left(\frac{y}{|y|}\right)\bigg|\bigg|\frac{\partial}{\partial y_i}\left(\frac{y_j}{|y|}\right)\bigg| \end{align} but \begin{align} \bigg|\frac{\partial}{\partial y_i}\left(\frac{y_j}{|y|}\right)\bigg| &= \bigg| \frac{\delta_{ij}}{|y|} - \frac{y_iy_j}{|y|^2}\bigg|\le \frac{\delta_{ij}}{|y|} + \frac{1}{2}\frac{y_i^2 + y_j^2}{|y|^2} \le \frac{1}{|y|} + C \end{align} This eventually leads to $$|\nabla_y \Omega (y/|y|)| \le C\sup_{z}|\nabla_z \Omega(z)|\left(1 + \frac{1}{|y|}\right).$$ How did he get this $1/|y|$ as a bound ? I don't get it.