Is there any injective group homomorphism from $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ to the group of units of $\mathbb{C}[x]/\langle {x^2}\rangle$?
My approach: For any such homomorphism $\varphi$, we will have $$\varphi((1,0))^2=\varphi(2(1,0))=\varphi((0,0))=1.$$ Thus $\varphi((1,0))=\pm 1$. WLOG if we assume $\varphi((1,0))=1$ then we must have $\varphi((0,1))=-1$, which will contradict the injectivity of $\varphi$ since $$\varphi((1,1))=\varphi((0,1))\varphi((1,0))=-1.$$
I suppose you meant injective group morphisms $\varphi:\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\longrightarrow\left(\mathbb{C}[x]/\langle x^2\rangle\right)^{\times}$. First, it will be helpful to know the units of the ring $\mathbb{C}[x]/\langle x^2\rangle$. It is clear that every element $p(x)\in\mathbb{C}[x]/\langle x^2\rangle$ is of the form $ax+b$ with $a,b\in\mathbb{C}$. Fixed $p(x)=ax+b$, we search for $g(x)=cx+d$ such that $p(x)g(x)=1$, then $$1=(ax+b)(cx+d)=acx^2+(ad+bc)x+bd=(ad+bc)x+bd.$$ Therefore $ad+bc=0$ and $bd=1$. Since $a$ and $b$ are fixed, that means that if $b=0$, $p(x)$ can not be an unit, and if $b\neq0$, then $d=\frac{1}{b}$ and $c=-\frac{ad}{b}=-\frac{a}{b^2}$. Therefore, any element as before with $b\neq0$ is an unit. Now, the commutativity is not a problem, so in order to have an injective morphism as wantes, we need two different unit elements of order $2$. By the previous calculations and using the same notation, that means that $a=-\frac{a}{b^2}$ and $b=\frac{1}{b}$. Notice that that means that $b^2=1$, so $b\in\{\pm1\}$. Now, if $a\neq0$, $b^2=-1$, and this is absurd, so $a=0$. We conclude that the only two elements of $\mathbb{C}[x]/\langle x^2\rangle$ of order $2$ are $\pm1$, so there can not be any injective group morphism as needed, since there is only one non trivial element in the group of order $2$.
In case you have trouble with the answer or any questions, do not hesitate to ask me.