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Rudin PMA 9.28. Why is $\phi(y)'k =(g(y)' k, k)$?

I am revising The implicit function theorem from Rudin's PMA as I forgot the entire proof of this theorem and I couldn't remember or understand why $\phi(y)'k =(g(y)' k, k)$ I didn't encounter this problem in the last time I read this theorem .

Here is the theorem (apologies for not writing the theorem but this would take too long to write):

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