Scaling dimension/weight of $\partial^{\mu}$ Under a scale transformation $x^{\mu}\rightarrow x^{'\mu}=\lambda x^{\mu}$. The operator $\partial_{\mu}$ has the conformal weight $1$ as $\partial_{\mu}^{'}=\frac{1}{\lambda}\partial_{\mu}$. I'm having confusion on the conformal weight of $\partial^{\mu}$. I tried to do it in following manner:
$$\partial^{'\mu}=g'^{\mu\nu}\partial_{\nu}^{'}$$
$$=(\lambda^{-2}g^{\mu\nu})(\frac{1}{\lambda}\partial_{\nu})$$
$$=\lambda^{-3}\partial^{\mu}$$
where in second line I used $g'_{\mu\nu}(x')=\lambda^2 g_{\mu\nu}(x)$ and $(\lambda A)^{-1}=\lambda^{-1} A^{-1}$ , for any matrix $A$ and any scalar $\lambda$.
This calculation seems to imply the weight of the operator $\partial^{\mu}$ to be $3$. But this is wrong since according to $(2.124)$ of CFT by Francesco the weight of $\partial^{\mu}$ should be same as $\partial_{\mu}$ i.e. $1$.
 A: Some explanations are better than others, sometimes even better than others in the same section of the same book. That's the case here. The question is easily resolved by focusing on the authors' better explanation of what they're doing. Here's a copy of their better explanation, from pages 36-37:

Consider a collection of fields, which we collectively denote by $\Phi$. The action functional will depend in general on $\Phi$ and its first derivatives:
$$
 S=\int d^dx\ \mathcal{L}(\Phi,\partial_\mu\Phi).
\tag{2.112}
$$
...
The change of the action functional under the transformation (2.113) [not copied] is obtained by substituting the new function $\Phi'(x)$ for the function $\Phi(x)$ (we note that the argument $x$ is the same in both cases). In other words, the new action is
$$
 S'=\int d^dx\ \mathcal{L}(\Phi'(x),\partial_\mu\Phi'(x)).
\tag{2.115}
$$

Notice that the equations can also be written like this:
$$
 S'[\Phi]=S[\Phi']
\tag{2.115$^*$}
$$
with
$$
 S[\Phi]\equiv \int d^dx\ \mathcal{L}(\Phi,\partial_\mu\Phi).
\tag{2.112$^*$}
$$
This shows clearly that the authors are transforming the fields, and only the fields. This is exactly as it should be.
Elsewhere in the same section, the authors describe the transformation as affecting both $\Phi$ and $x$. That's unnessary (and confusing IMO), because in the action, the spacetime argument $x$ is just a dummy integration variable. Conceptually, it plays the same role as the dummy index in a sum. The action $S$ does not depend on $x$, just like the sum $\sum_n c_n$ does not depend on $n$. The sum depends on the quantities $c_n$ that are indexed by $n$, and the action $S$ depends on the fields $\Phi$ that are indexed by $x$.
The action does depend on the background metric $g_{\mu\nu}$, but the authors are not transforming the metric. The metric is regarded as a prescribed background, part of the model's definition. That's why the authors only transform the fields $\Phi$, not the metric.
With those clarifications, the answer to the question is straightforward. Consider the action
$$
 S[\Phi]=\int d^dx\ g^{\mu\nu}(\partial_\mu\Phi)(\partial_\nu\Phi)
\tag{2.123}
$$
with fixed and $x$-independent coefficients $g^{\mu\nu}$. The only input to the action that isn't fixed is $\Phi$. If we define a new input $\Phi'$ by the condition
$$
 \Phi'(\lambda x)\equiv \lambda^{-\Delta}\Phi(x),
\tag{2.121}
$$
which can also be written
$$
 \Phi'(x)\equiv \lambda^{-\Delta}\Phi(\lambda^{-1} x),
\tag{2.121$^*$}
$$
then we find that the action is invariant (that is, $S[\Phi']=S[\Phi]$) if and only if
$$
 \Delta = \frac{d}{2}-1,
\tag{2.124}
$$
exactly as the book says. The proof involves an ordinary change of integration variable, which is just plain calculus. The language about the "weight" of $\partial_\mu$ is just a way of saying what happens to $\partial_\mu$ as a result of this ordinary change of integration variable, which explains why the "weight" of $\partial^\mu$ is the same.
