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Use fseek to find the size of file and then read backwards from the end until n line breaks are read and then print the result.
Another solution can be like this:
Take a buffer for n lines and keep updating the buffer while reading from the files. So that at last the buffer contains the last n lines.

You can use a doubly linked list with a fixed size of N.
InsertLast and RemoveFirst.

In the end, you'll have the last N lines.

Use a Queue with a size limit of n, whenever n is exceed dequeue. Do this through the whole file if reading forward is the only option available.

If there is a way to read the file stream in reverse, then that's the best option.

We can read a file data in vector and then read n lines from vector in reverse order.

Store N lines in a vector. When N lines have been read, start adding to the beginning again and increment a counter by 1 % size of the array for each element read.

readNLines() {
	int counter = 0;
	while(file NOT empty) {
		if(vector.size() < n)
			vector.push_back(file.line)
		else {
			vector[(counter+1)%n] = file.line
			counter = (counter+1)%n
		}
	}
	for(int i = 1; (counter+i)%n != counter; ++i) {
	output vector[i-1]
	}
}

#include <iostream>
#include <queue>
#include <fstream>

using namespace std;
void printLastNLines(const std::string& fname, int n);

int main()
{
    std::string fname = "myfile.txt";
    int n=10;
    printLastNLines(fname, n);
    return 0;
}

void printLastNLines(const std::string& fname, int n)
{
    ifstream infile(fname.c_str());
    if (!infile.is_open())
    {
        cout<<"File opening error."<<endl;
        return;
    }
    std::queue<std::string> lines;
    std::string line;
    while (std::getline(infile, line))
    {
        if (lines.size() >= n)
        {
            lines.pop();
        }
        lines.push(line);
    }
    infile.close();

    while(!lines.empty())
    {
        cout<<"line = "<<lines.front()<<endl;
        lines.pop();
    }
}

Go through all the lines one by one and put each line in a node of a single linked list.
Then, take two pointers, one starting at head, and other starting at a distance of n from head

At the end of one traversal of the linked list, you have the first pointer at a distance of n from the end and second pointer at the end.

Just print the lines