Find $\lambda\in\mathbb{R}$ such that $y=e^x$ and $y=\lambda x^2$ touch. 
Find $\lambda\in\mathbb{R}$ such that $y=e^x$ and $y=\lambda x^2$ touch.

I'm just a beginner on derivatives, and I guess it should be done using them, but I'm totally stuck.
(I guess it's not "touch" in English but oh well.)
 A: Let me translate this problem:
You want to find a $\lambda \in \mathbb{R}$ so that there exists a point $a\in \mathbb{R}$ with the property:
$f(a)=g(a)$ (the curves intersect) and $f'(a)=g'(a)$ (their tangents are parallel match exactly) where $f(x)=e^x$ and $g(x)=\lambda x^2$. Can you do this? 
Hint: $f'(x)=e^x=f(x)$ which means you want $g'(a)=g(a)$. Solve this for $\lambda,a$ and check it against the condition $f(a)=g(a)$
A: Let $f(x)=e^x$ and $g(x)=\lambda x^2$. Then $f(x)$ and $g(x)$ are tangent if, at some point $x_0$:
$$f(x_0)=g(x_0)$$
(the curves meet) 
and:
$$f'(x_0)=g'(x_0)$$
(the curves have the same slope where they meet.)
What is $f'(x)$? What is $g'(x)$? This gives you two equations in two variables, $x_0$ and $\lambda$. Solve for $x_0$ first, and then solve for $\lambda$.
A: $y=e^{x}$ and $y=\lambda x^{2}$ are tangent at $x_{0}$ iff the curves and their derivatives are equal at $x_{0}$.  
Thus, $e^{x_{0}}=\lambda x_{0}^{2}$ and $e^{x_{0}}=2\lambda x_{0}$. Therefore, $\lambda x_{0}^{2}=2\lambda x_{0} \implies x_{0}=2$. Substituting gives $\lambda=e^{2}/4$.
You can see this using graphing software.
