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I am studying the inverse function theorem in Differential Topology which says that if $f$ is a smooth map between two manifolds $X$ and $Y$, whose derivative at the point $x$ is an isomorphism, then $f$ is a local diffeomorphism at $x$.

We know that the derivative is a linear map, but an isomophism must be a bijection and differentiation is not a bijection. For example $x^2 + 5$ and $x^2 + 3$ go to the same derivative, namely $2x$. May you give me an example in which this theorem is applied, please?

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You're misinterpreting what "bijection" refers to here. It doesn't mean "Differentiation is a bijection from smooth maps to linear maps", but rather "The linear map you get when you differentiate your given map at $x$ is a bijection". And for any given $x \neq 0$, the linear map given by multiplication by $2x$ is certainly a bijection from $\Bbb R$ to $\Bbb R$. This means that around any point except $x = 0$, $x^2 + 3$ and $x^2 + 5$ are local diffeomorphisms.

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    Thank you for your reply. I am really confused in the fact that the derivative of a function at a specific point x is a number or a matrix. In this case, the derivative of a function f at a point x is a mapping from the tangent space of the manifold X at the point x to the tangent space of the manifold Y at the point y = f (x). I have difficulties in concepting this fact. How can it be possible? – 2017-01-08
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    Why would it not be possible? They are both finite dimensional vector spaces. Have you learned what the tangent space is, how the differential is calculated, and why it is calculated that way? – 2017-01-09
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    @User1999 In this case, I like to think about the tangent space as equivalence classes of curves going through the point (where two curves are equivalent if they go through the point in the same direction and speed). That way, it's quite clear that a curve going through $x$ gets sent by $f$ to a curve going though $y$. The derivative of $f$ is simply a correspondence saying which equivalence classes of curves through $x$ gets sent to which equivalence classes of curves through $y$. – 2017-01-09
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    (cont.) If that correspondence is an isomorphism, that means that the area around $f(x)\in f(X)$ (i.e. the image of a neighbourhood of $x$) and around $y \in Y$ (i.e. a neighbourhood of $y$) looks alike, since curves can go through the two points from as many directions. In my example, if $x = 0$, then in $f(X)$ around $f(x)$, curves can only come from above and exit upwards, which means that it's not a local diffeomorphism there. Actually, any curve through $x = 0$ gets sent to the $0$ class in $TY_{f(x)}$ (any curve has to turn around in a smooth way, which means it must be stationary). – 2017-01-09