Confused about partial derivatives I am having some issues understanding what should I keep constant and what not in certain cases when I take partial derivatives. Specifically in this kind of situation: say we have a function
$$f(x,y) = x^3+7y^2$$
and we also know that $y=2x+1$ and we need to find the partial derivative with respect to $x$ at a given point, say $\frac{\partial f(2,3)}{\partial x}$. From what I understood, when taking partial derivatives with respect to a variable, you need to keep the other constant, in which case, if I do that above (keeping $y$ constant) I would get, $\frac{\partial f(x,y)}{\partial x}=3x^2$, so I get $12$. However, if I plug $x$ in $y$ explicitly I would get $$f(x,y) = x^3+7(2x+1)^2$$ so,
$$\frac{\partial f(x,y)}{\partial x}=3x^2+28(2x+1)$$
So, I get $152$. What should I do? Thank you!

It's definitely important to make some distinctions with notation. Given a function ${f(x,y)}$, then ${\frac{\partial f}{\partial x}}$ means "treating $y$ as a constant, what is the derivative of $f$ with respect to ${x}$?" <- this does not, on it's own, necessary tell you about how $f$ changes overall with respect to $x$. In order to know this, we must use the multi-variable chain rule, which is:
$$
\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}
$$
notice if ${\frac{dy}{dx}=0}$, i.e. ${x}$ and ${y}$ are independent, then ${\frac{df}{dx}=\frac{\partial f}{\partial x}}$ (i.e. the partial derivative tells us all we need to know about how $f$ changes with respect to $x$).
