Let $x'=x$ and $y'=y-ax$ for some nonzero constant $a$. Then the chain rule states that
$\frac{\partial}{\partial x}=\frac{\partial y'}{\partial x}\frac{\partial}{\partial y'}+\frac{\partial x'}{\partial x}\frac{\partial}{\partial x'}= -a\frac{\partial}{\partial y'}+\frac{\partial}{\partial x'}$.
Here I have a serious confusion. Since $x'=x$, isn't it that $\frac{\partial}{\partial y'}=0$? However this conclusion is absurd because $\frac{\partial y'}{\partial y'}=1$ What is wrong? I am extremely confused....
When you are learning chain rule, it's better to be explicit about which functions you are applying it to. Let's let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ and $g_{i}:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable functions for each $i=1,\ldots,n$. Define the function $h$ by $$ h(x)=f(g_{1}(x),\ldots,g_{n}(x)). $$ Then, the chain rule tells us how to get the derivative of $h$: $$ h^{\prime}(x)=\frac{\partial f}{\partial x_{1}}(g_{1}(x))g_{1}^{\prime}(x)+\cdots+\frac{\partial f}{\partial x_{n}}(g_{n}(x))g_{n}^{\prime}(x). $$ In the above, we are using $\frac{\partial f}{\partial x_{i}}$ to denote the derivative of $f$ with respect to its $i$-th argument.
Example: Let $f(x_{1},x_{2})=(x_{1})^{3}+(x_{2})^{2}$. Let $g_{1}(x)=2x$ and $g_{2}(x)=\sin x$. Then, $$ h(x)=f(g_{1}(x),g_{2}(x))=(2x)^{3}+(\sin x)^{2}. $$ The chain rule tell us that $$ h^{\prime}(x)=\underbrace{\frac{\partial f}{\partial x_{1}}(g_{1}(x))}_{3(2x)^{2}}\underbrace{g_{1}^{\prime}(x)}_{2}+\underbrace{\frac{\partial f}{\partial x_{2}}(g_{2}(x))}_{2(\sin x)}\underbrace{g_{2}^{\prime}(x)}_{\cos x}. $$